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* These questions are only for students studying for the Optometry Admission Test (OAT). If you are taking the DAT, this section is not applicable to you. There is no physics on the DAT.
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Question 1 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
1. Question
1. Elena rides her bike 30 m/s at 60^{o} North of East. What are the x and y components of her velocity?Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Velocity (v) = 30 m/s
 θ = 60^{o}
Solving for: x and y components of her velocity (v_{x} and v_{y})
Step 2. Next, let’s work backwards. What do we need in order to solve for the x and y components of her velocity (v_{x} and v_{y})?
For many different types of problems on the OAT, you will need to know that:1. v_{x} = v cos (θ)
2. v_{y} = v sin (θ)
where “v” is any vector. These are based on trigonometry. If you’re not already, get very familiar with these equations.Step 3. We know all the variables we need to solve for V_{x} and V_{y}. Plug them in to get your answer.
Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Velocity (v) = 30 m/s
 θ = 60^{o}
Solving for: x and y components of her velocity (v_{x} and v_{y})
Step 2. Next, let’s work backwards. What do we need in order to solve for the x and y components of her velocity (v_{x} and v_{y})?
For many different types of problems on the OAT, you will need to know that:1. v_{x} = v cos (θ)
2. v_{y} = v sin (θ)
where “v” is any vector. These are based on trigonometry. If you’re not already, get very familiar with these equations.Step 3. We know all the variables we need to solve for V_{x} and V_{y}. Plug them in to get your answer.
Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 2 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
2. Question
2. The resultant vector of the following vectors (A + B) points in which direction?
Correct / You marked this questionStep 1. First, let’s write down our variables and what we are solving for. We have already been provided with a sketch.
 A = √2
 θ_{A} = 60^{o}
 B = 1
 θ_{B} = 45^{o}
Solving for: Direction of the resultant vector (A+B)
Step 2. Next, let’s work backwards. What do we need in order to solve for the direction of the resultant vector (A+B)?
The resultant vector (let’s call this “C”) will have: x component (C_{x}) that equals the sum of the x component of A (A_{x}) and the x component of B (B_{x})
 y component (C_{y}) that equals the sum of the y component of A (A_{y}) and the y component of B (B_{y})
Once we determine C_{x} and C_{y}, we can determine the direction of C.
Step 3. Next, solve for C_{x} and C_{y}
To solve for C_{x} , we need to solve for A_{x} and B_{x}Note that A_{x} and B_{x} point in opposite directions. Convention dictates we define towards the right as positive, and to the left as negative. Two vectors of equal magnitude pointing in exactly opposite directions cancel each other out.
Thus, C_{x} = 0. We can immediately cross out any answer options which have an x component (i.e. any arrows not pointing directly up or down). Thus, C = C_{y}
To solve for C_{y} , we need to solve for A_{y} and B_{y}
Note that A_{y} and B_{y} point in opposite directions. Convention dictates we define up as positive, and down negative. The magnitude of A is greater than the magnitude of B. Thus, C_{y} (and also C) points up ↑
Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1. First, let’s write down our variables and what we are solving for. We have already been provided with a sketch.
 A = √2
 θ_{A} = 60^{o}
 B = 1
 θ_{B} = 45^{o}
Solving for: Direction of the resultant vector (A+B)
Step 2. Next, let’s work backwards. What do we need in order to solve for the direction of the resultant vector (A+B)?
The resultant vector (let’s call this “C”) will have: x component (C_{x}) that equals the sum of the x component of A (A_{x}) and the x component of B (B_{x})
 y component (C_{y}) that equals the sum of the y component of A (A_{y}) and the y component of B (B_{y})
Once we determine C_{x} and C_{y}, we can determine the direction of C.
Step 3. Next, solve for C_{x} and C_{y}
To solve for C_{x} , we need to solve for A_{x} and B_{x}Note that A_{x} and B_{x} point in opposite directions. Convention dictates we define towards the right as positive, and to the left as negative. Two vectors of equal magnitude pointing in exactly opposite directions cancel each other out.
Thus, C_{x} = 0. We can immediately cross out any answer options which have an x component (i.e. any arrows not pointing directly up or down). Thus, C = C_{y}
To solve for C_{y} , we need to solve for A_{y} and B_{y}
Note that A_{y} and B_{y} point in opposite directions. Convention dictates we define up as positive, and down negative. The magnitude of A is greater than the magnitude of B. Thus, C_{y} (and also C) points up ↑
Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 3 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
3. Question
3. Kathryn runs 30 meters to the South to her friends house, then runs 40 meters East to school. What is the best way to describe her displacement vector?Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Horizontal displacement (D_{x}) = 30 meters
 Vertical displacement(D_{y}) = 40 meters
Solving for: Net displacement vector (D)
Step 2. Next, let’s work backwards. What do we need in order to solve for the net displacement vector (D)?
We already are given the x and y components of the girl’s displacement. Thus, we can use the pythagorean equation to solve for D (i.e. the hypotenuse).Step 3. Plug D_{x} and D_{y} into the pythagorean equation to solve for D.
Pythagorean Equation: x^{2} + y^{2} = z^{2}(D_{x})^{2} + (D_{y})^{2} = (D)^{2}
(30m)^{2} + (40m)^{2} = (D)^{2}
900m^{2} + 1600m^{2} = (D)^{2}
2500m^{2} = (D)^{2}
D =√2500m^{2} = 50 metersRemember, we are asked to describe the displacement vector! A vector has both magnitude and a direction.
Thus, as you can see from our sketch, vector D = 50 meters Southeast, our answer is A
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Horizontal displacement (D_{x}) = 30 meters
 Vertical displacement(D_{y}) = 40 meters
Solving for: Net displacement vector (D)
Step 2. Next, let’s work backwards. What do we need in order to solve for the net displacement vector (D)?
We already are given the x and y components of the girl’s displacement. Thus, we can use the pythagorean equation to solve for D (i.e. the hypotenuse).Step 3. Plug D_{x} and D_{y} into the pythagorean equation to solve for D.
Pythagorean Equation: x^{2} + y^{2} = z^{2}(D_{x})^{2} + (D_{y})^{2} = (D)^{2}
(30m)^{2} + (40m)^{2} = (D)^{2}
900m^{2} + 1600m^{2} = (D)^{2}
2500m^{2} = (D)^{2}
D =√2500m^{2} = 50 metersRemember, we are asked to describe the displacement vector! A vector has both magnitude and a direction.
Thus, as you can see from our sketch, vector D = 50 meters Southeast, our answer is A

Question 4 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
4. Question
4. Which of the following is NOT a vector quantity?Correct / You marked this questionStep 1. First, let’s recall what exactly a vector quantity is:
 A vector quantity is a quantity (i.e. a numerical amount) that has BOTH a magnitude AND a direction.
 A scalar quantity is a quantity (i.e. a numerical amount) that has ONLY a magnitude.
Step 2. Let’s tackle our options one by one to determine whether they are vector quantities.
[A] Displacement: displacement is the change in position of an object. Displacement IS a vector quantity. If you think about it, it wouldn’t make sense for displacement to be a scalar quantity. If you said “the bike rider’s displacement is 30 meters” without including in what direction, that wouldn’t tell you much about the final position of the biker. Distance is the scalar component of displacement.[B] Force: Force is a push or pull on an object. Force IS a vector quantity. If you think about it, it wouldn’t make sense for force to be a scalar quantity: If you want to know the net force on a box from two people pushing on it, it wouldn’t be possible unless you knew in what direction they were pushing.
[C] Mass: Mass is the amount of matter in an object, which dictates the object’s inertia. Mass is NOT a vector quantity. Mass is a scalar quantity. Mass and weight are often confused. An object’s mass is constant, regardless of location. An object’s weight is a vector quantity, and depends on its gravitational attraction towards (in the direction of!) another body.
[D] Velocity: Velocity is an object’s displacement per unit time. Velocity IS a vector quantity. The scalar equivalent of velocity is speed. Speed tells you how fast an object is moving, while velocity tells you both the object’s speed and the direction it is moving in.
[E] Weight: Weight is an object’s gravitational attraction towards another body. Weight IS a vector quantity. Weight is a type of force, with units of newtons. Think about it: your weight is the pull you feel towards (in the direction of!) the Earth.
Incorrect / You marked this questionStep 1. First, let’s recall what exactly a vector quantity is:
 A vector quantity is a quantity (i.e. a numerical amount) that has BOTH a magnitude AND a direction.
 A scalar quantity is a quantity (i.e. a numerical amount) that has ONLY a magnitude.
Step 2. Let’s tackle our options one by one to determine whether they are vector quantities.
[A] Displacement: displacement is the change in position of an object. Displacement IS a vector quantity. If you think about it, it wouldn’t make sense for displacement to be a scalar quantity. If you said “the bike rider’s displacement is 30 meters” without including in what direction, that wouldn’t tell you much about the final position of the biker. Distance is the scalar component of displacement.[B] Force: Force is a push or pull on an object. Force IS a vector quantity. If you think about it, it wouldn’t make sense for force to be a scalar quantity: If you want to know the net force on a box from two people pushing on it, it wouldn’t be possible unless you knew in what direction they were pushing.
[C] Mass: Mass is the amount of matter in an object, which dictates the object’s inertia. Mass is NOT a vector quantity. Mass is a scalar quantity. Mass and weight are often confused. An object’s mass is constant, regardless of location. An object’s weight is a vector quantity, and depends on its gravitational attraction towards (in the direction of!) another body.
[D] Velocity: Velocity is an object’s displacement per unit time. Velocity IS a vector quantity. The scalar equivalent of velocity is speed. Speed tells you how fast an object is moving, while velocity tells you both the object’s speed and the direction it is moving in.
[E] Weight: Weight is an object’s gravitational attraction towards another body. Weight IS a vector quantity. Weight is a type of force, with units of newtons. Think about it: your weight is the pull you feel towards (in the direction of!) the Earth.

Question 5 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
5. Question
5. Timmy pulls his 10 kg wagon with 10N of force at 30^{o} above the horizontal. What are the x and y components of this force vector?Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Force (F) = 10 N
 θ = 30^{o}
 Mass = 10 kg
Solving For: x and y components of Force vector (F_{x} and F_{y} )
Step 2. Next, let’s work backwards. What do we need in order to solve for the x and y components of his velocity (F_{x} and F_{y} )?
Based on trigonometry, we know that:1. F_{x} = F cos (θ)
2. F_{y} = F sin (θ)Note: we do not need the wagon’s mass to be able to solve for F_{x} and F_{y} . This variable was included to confuse you.
Step 3. We know all the variables we need to solve for F_{x} and F_{y} . Plug them in to get your answer.
Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Force (F) = 10 N
 θ = 30^{o}
 Mass = 10 kg
Solving For: x and y components of Force vector (F_{x} and F_{y} )
Step 2. Next, let’s work backwards. What do we need in order to solve for the x and y components of his velocity (F_{x} and F_{y} )?
Based on trigonometry, we know that:1. F_{x} = F cos (θ)
2. F_{y} = F sin (θ)Note: we do not need the wagon’s mass to be able to solve for F_{x} and F_{y} . This variable was included to confuse you.
Step 3. We know all the variables we need to solve for F_{x} and F_{y} . Plug them in to get your answer.
Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 6 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
6. Question
6. Mark is on a run listening to music on his phone. He runs 20 meters East, then realizes he dropped his phone. He turns and runs back 15 meters West to grab it. He starts running North, and gets 30 meters before he realizes he dropped his phone again! He runs back 18 meters South to grab it. What is the magnitude of Mark’s displacement?Correct / You marked this questionStep 1. First, let’s write down a quick sketch, and what we are solving for:
Solving for: Mark’s displacementStep 2. Next, let’s work backwards. What do we need in order to solve for the magnitude of Mark’s displacement?
To determine the magnitude of Mark’s displacement, we need to calculate both the x and y components of his displacement.In the x direction, Mark ran 20 meters East (+20m) and then 15 meters West ( 15m). His net displacement in the x direction is +20m – 15m = 5 m
In the y direction, Mark ran 30 meters North (+ 30m) and then 18 meters South ( 18m). His net displacement in the y direction is +30m – 18m = 12 m
Redrawing our picture, we now have:
Step 3. This has Pythagorean equation written all over it! Plug in our x and y components to solve for Displacement (the hypotenuse).
Pythagorean Equation: x^{2} + y^{2} = z^{2}(D_{x})^{2} + (D_{y})^{2} = (D)^{2}
(5m)^{2} + (12m)^{2} = (D)^{2}
25m^{2} + 144m^{2} = (D)^{2}
169m^{2} = (D)^{2}
D = √169m^{2} = 13 metersIncorrect / You marked this questionStep 1. First, let’s write down a quick sketch, and what we are solving for:
Solving for: Mark’s displacementStep 2. Next, let’s work backwards. What do we need in order to solve for the magnitude of Mark’s displacement?
To determine the magnitude of Mark’s displacement, we need to calculate both the x and y components of his displacement.In the x direction, Mark ran 20 meters East (+20m) and then 15 meters West ( 15m). His net displacement in the x direction is +20m – 15m = 5 m
In the y direction, Mark ran 30 meters North (+ 30m) and then 18 meters South ( 18m). His net displacement in the y direction is +30m – 18m = 12 m
Redrawing our picture, we now have:
Step 3. This has Pythagorean equation written all over it! Plug in our x and y components to solve for Displacement (the hypotenuse).
Pythagorean Equation: x^{2} + y^{2} = z^{2}(D_{x})^{2} + (D_{y})^{2} = (D)^{2}
(5m)^{2} + (12m)^{2} = (D)^{2}
25m^{2} + 144m^{2} = (D)^{2}
169m^{2} = (D)^{2}
D = √169m^{2} = 13 meters 
Question 7 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
7. Question
7. What are the magnitudes of the x and y components of the resultant vector (A+B)?
Correct / You marked this questionStep 1. First, let’s write down our variables, and what we are solving for. We have already been provided with a sketch.
 A = 4 √3
 θ_{A} = 30^{o}
 B = 2 √3
 θ_{B} = 30^{o}
Solving for: Magnitudes of the x and y components of the resultant vector (A+B)
Step 2. Next, let’s work backwards. What do we need in order to solve for the magnitudes of the x and y components of the resultant vector (A+B)?
The resultant vector (let’s call this “C”) will have: x component (C_{x}) that equals the sum of the x component of A (A_{x}) and the x component of B (B_{x})
 y component (C_{y}) that equals the sum of the y component of A (A_{y}) and the y component of B (B_{y})
Step 3. Next, solve for C_{x} and C_{y}
To solve for C_{x} , we need to solve for A_{x} and B_{x}Note that A_{x} and B_{x} point in the same direction. Convention dictates we define towards the right as positive, and to the left as negative. To get the net magnitude, we add the two magnitudes together:
C_{x} = A_{x} + B_{x} = 6 + 3 = 9
To solve for C_{y} , we need to solve for A_{y} and B_{y}
Note that A_{y} and B_{y} point in opposite directions. Convention dictates we define up as positive, and down negative. To get the net magnitude, we add the two magnitudes together:
Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1. First, let’s write down our variables, and what we are solving for. We have already been provided with a sketch.
 A = 4 √3
 θ_{A} = 30^{o}
 B = 2 √3
 θ_{B} = 30^{o}
Solving for: Magnitudes of the x and y components of the resultant vector (A+B)
Step 2. Next, let’s work backwards. What do we need in order to solve for the magnitudes of the x and y components of the resultant vector (A+B)?
The resultant vector (let’s call this “C”) will have: x component (C_{x}) that equals the sum of the x component of A (A_{x}) and the x component of B (B_{x})
 y component (C_{y}) that equals the sum of the y component of A (A_{y}) and the y component of B (B_{y})
Step 3. Next, solve for C_{x} and C_{y}
To solve for C_{x} , we need to solve for A_{x} and B_{x}Note that A_{x} and B_{x} point in the same direction. Convention dictates we define towards the right as positive, and to the left as negative. To get the net magnitude, we add the two magnitudes together:
C_{x} = A_{x} + B_{x} = 6 + 3 = 9
To solve for C_{y} , we need to solve for A_{y} and B_{y}
Note that A_{y} and B_{y} point in opposite directions. Convention dictates we define up as positive, and down negative. To get the net magnitude, we add the two magnitudes together:
Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 8 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
8. Question
8. Three friends are moving a couch. Howard and Kelsey are pushing the couch in the same direction, each with a force of 15 Newtons. Zack is pushing perpendicular to Howard and Kelsey with a force of 40 Newtons. What is the magnitude of the net force on the couch?Correct / You marked this questionStep 1. First, let’s write down a quick sketch, and what we are solving for:
 F_{1} (Howard) = 15 N
 F_{2} (Kelsey) = 15 N
 F_{3} (Zack) = 40 N
Solving for: the magnitude of the net force on the couch.
Step 2. Next, let’s work backwards. What do we need in order to solve for the magnitude of the net force on the couch?
Note: We were given that Zack’s force was perpendicular to Howard and Kelsey’s. There are several different ways we could have drawn our sketch, but they would all yield a resultant net force vector with the same magnitude. The rest of the solution will refer to the way we drew our sketch above.To determine the magnitude of the net force on the couch, we need to calculate both the x and y components of the net force on the couch.
In the x direction, Kelsey pushed with 15 Newtons of force and Howard pushed with 15 Newtons of force, both to the right. 15N + 15N = +30N
In the y direction, Zack pushed with 40 Newtons downward = 40N
Redrawing our picture, we now have:
Step 3. Let’s refer to our old friend Pythagoras for this one. Plug in our x and y components to solve for the net force vector (the hypotenuse).
Pythagorean Equation: x^{2} + y^{2} = z^{2}(F_{x})^{2} + (F_{y})^{2} = (F)^{2}
(30N)^{2} + (40N)^{2} = (F)^{2}
900N^{2} + 1600N^{2} = (F)^{2}
2500m^{2} = (F)^{2}
F = √2500 N^{2} = 50 NRemember, this question only asks for the magnitude of the net force. We are NOT required to provide any information about its direction. Even if we did want to describe its direction, the information provided in the question is insufficient to solve for this.
Incorrect / You marked this questionStep 1. First, let’s write down a quick sketch, and what we are solving for:
 F_{1} (Howard) = 15 N
 F_{2} (Kelsey) = 15 N
 F_{3} (Zack) = 40 N
Solving for: the magnitude of the net force on the couch.
Step 2. Next, let’s work backwards. What do we need in order to solve for the magnitude of the net force on the couch?
Note: We were given that Zack’s force was perpendicular to Howard and Kelsey’s. There are several different ways we could have drawn our sketch, but they would all yield a resultant net force vector with the same magnitude. The rest of the solution will refer to the way we drew our sketch above.To determine the magnitude of the net force on the couch, we need to calculate both the x and y components of the net force on the couch.
In the x direction, Kelsey pushed with 15 Newtons of force and Howard pushed with 15 Newtons of force, both to the right. 15N + 15N = +30N
In the y direction, Zack pushed with 40 Newtons downward = 40N
Redrawing our picture, we now have:
Step 3. Let’s refer to our old friend Pythagoras for this one. Plug in our x and y components to solve for the net force vector (the hypotenuse).
Pythagorean Equation: x^{2} + y^{2} = z^{2}(F_{x})^{2} + (F_{y})^{2} = (F)^{2}
(30N)^{2} + (40N)^{2} = (F)^{2}
900N^{2} + 1600N^{2} = (F)^{2}
2500m^{2} = (F)^{2}
F = √2500 N^{2} = 50 NRemember, this question only asks for the magnitude of the net force. We are NOT required to provide any information about its direction. Even if we did want to describe its direction, the information provided in the question is insufficient to solve for this.

Question 9 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
9. Question
9. A scalar quantity has:Correct / You marked this questionStep 1. This question is based purely on knowing the definition of a scalar quantity.
 A scalar quantity is a quantity (i.e. a numerical amount) that has ONLY a magnitude.
 For context, a vector quantity is a quantity (i.e. a numerical amount) that has BOTH a magnitude AND a direction.
Step 2. To be thorough, let’s look at the answer options one by one:
[A] Both magnitude and direction: this is the definition of a vector.[B] Direction: a scalar quantity does NOT have direction. A vector quantity does.
[C] Magnitude: yes, a scalar quantity has a magnitude.
[D] Weight: weight is a vector quantity. Don’t get confused by “scale.”
[E] Only Integer Values: this is incorrect. A scalar magnitude could be any numerical value, not necessarily limited to integers.
Incorrect / You marked this questionStep 1. This question is based purely on knowing the definition of a scalar quantity.
 A scalar quantity is a quantity (i.e. a numerical amount) that has ONLY a magnitude.
 For context, a vector quantity is a quantity (i.e. a numerical amount) that has BOTH a magnitude AND a direction.
Step 2. To be thorough, let’s look at the answer options one by one:
[A] Both magnitude and direction: this is the definition of a vector.[B] Direction: a scalar quantity does NOT have direction. A vector quantity does.
[C] Magnitude: yes, a scalar quantity has a magnitude.
[D] Weight: weight is a vector quantity. Don’t get confused by “scale.”
[E] Only Integer Values: this is incorrect. A scalar magnitude could be any numerical value, not necessarily limited to integers.

Question 10 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
10. Question
10. The resultant vector of the following vectors points in which direction?
Correct / You marked this questionStep 1. There isn’t much math / calculation involved in this problem. I would call this one that you can “eyeball” without listing out the variables to save time. You are already provided a sketch.
Solving for: Direction of the resultant vector (A+B+C+D+E)Step 2. Add up the individual force vectors to determine direction of combined net force vector:
 A and D both point right. Thus, their net force points right →
 B and C both have the same magnitude and point in exactly opposite directions. Their net force = 0
 E points up ↑
 The net force of (A & D) and E points both right → and up ↑. Thus, the direction of the resultant force vector of all vectors is
Incorrect / You marked this questionStep 1. There isn’t much math / calculation involved in this problem. I would call this one that you can “eyeball” without listing out the variables to save time. You are already provided a sketch.
Solving for: Direction of the resultant vector (A+B+C+D+E)Step 2. Add up the individual force vectors to determine direction of combined net force vector:
 A and D both point right. Thus, their net force points right →
 B and C both have the same magnitude and point in exactly opposite directions. Their net force = 0
 E points up ↑
 The net force of (A & D) and E points both right → and up ↑. Thus, the direction of the resultant force vector of all vectors is

Question 11 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
11. Question
11. Tron is riding on an electric grid. He starts out at (2,2). He rides to (4,2), then he rides to (4,6 ). What is Tron’s displacement?Correct / You marked this questionStep 1. First, let’s write down a quick sketch, and what we are solving for:
Solving for: Tron’s displacement
Step 2. This one should shout Pythagorean equation from the rooftops! First, let’s calculate D_{x} and D_{y} (the horizontal and vertical components of Tron’s displacement) to be able to plug in later.
D_{x} = 4 – (2) = 6
D_{y} = 6 – (2) = 8When in doubt, there’s no shame in counting the little boxes 🙂
Step 3. Plug D_{x} and D_{y} into the Pythagorean equation to solve for D.
Pythagorean Equation: x^{2} + y^{2} = z^{2}
(D_{x})^{2} + (D_{y})^{2} = (D)^{2}
(6)^{2} + (8)^{2} = (D)^{2}
36 + 64 = (D)^{2}
100 = (D)^{2}
D = √100 = 10Incorrect / You marked this questionStep 1. First, let’s write down a quick sketch, and what we are solving for:
Solving for: Tron’s displacement
Step 2. This one should shout Pythagorean equation from the rooftops! First, let’s calculate D_{x} and D_{y} (the horizontal and vertical components of Tron’s displacement) to be able to plug in later.
D_{x} = 4 – (2) = 6
D_{y} = 6 – (2) = 8When in doubt, there’s no shame in counting the little boxes 🙂
Step 3. Plug D_{x} and D_{y} into the Pythagorean equation to solve for D.
Pythagorean Equation: x^{2} + y^{2} = z^{2}
(D_{x})^{2} + (D_{y})^{2} = (D)^{2}
(6)^{2} + (8)^{2} = (D)^{2}
36 + 64 = (D)^{2}
100 = (D)^{2}
D = √100 = 10 
Question 12 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
12. Question
12. A 1 kg hockey puck is stopped on the ice rink. Two players from the Mighty Ducks slap the puck at exactly the same time, from opposite sides. Player A hits the puck with 10√3 N of force 30^{o} above the ice. Player B hits the puck with 30 N of force 60^{o} above the ice. What is the net force on the puck due to the players’ sticks?Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Force Player A (A) = 10√3 N
 θ = 30^{o}
 Force Player B (B) = 30 N
 θ = 60^{o}
Solving for: the net force on the puck due to the players’ sticks
Step 2. Next, let’s work backwards. What do we need in order to solve for the net force on the puck due to the players’ sticks?
Note: from the get go, you can discard the mass of the puck as irrelevant information. The question specifically asks for the net force on the puck due to the players’ sticks, not due to its own weight.
Moving on, the resultant vector (let’s call this “C”) will have:
 x component (C_{x}) that equals the sum of the x component of A (A_{x}) and the x component of B (B_{x})
 y component (C_{y}) that equals the sum of the y component of A (A_{y}) and the y component of B (B_{y})
Once we determine C_{x} and C_{y}, we can determine the direction and magnitude of C.
Step 3. Next, solve for C_{x} and C_{y}
To solve for C_{x}, we need to solve for A_{x} and B_{x}
Note that A_{x} and B_{x} point in opposite directions. Convention dictates we define towards the right as positive, and to the left as negative. Two vectors of equal magnitude pointing in exactly opposite directions cancel each other out.
Thus, C_{x} = 0. This implies that C = C_{y}. To solve for C_{y}, we need to solve for A_{y} and B_{y}
Note that A_{y} and B_{y} point in the same direction. Convention dictates we define up as positive, and down negative.
Thus,Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Force Player A (A) = 10√3 N
 θ = 30^{o}
 Force Player B (B) = 30 N
 θ = 60^{o}
Solving for: the net force on the puck due to the players’ sticks
Step 2. Next, let’s work backwards. What do we need in order to solve for the net force on the puck due to the players’ sticks?
Note: from the get go, you can discard the mass of the puck as irrelevant information. The question specifically asks for the net force on the puck due to the players’ sticks, not due to its own weight.
Moving on, the resultant vector (let’s call this “C”) will have:
 x component (C_{x}) that equals the sum of the x component of A (A_{x}) and the x component of B (B_{x})
 y component (C_{y}) that equals the sum of the y component of A (A_{y}) and the y component of B (B_{y})
Once we determine C_{x} and C_{y}, we can determine the direction and magnitude of C.
Step 3. Next, solve for C_{x} and C_{y}
To solve for C_{x}, we need to solve for A_{x} and B_{x}
Note that A_{x} and B_{x} point in opposite directions. Convention dictates we define towards the right as positive, and to the left as negative. Two vectors of equal magnitude pointing in exactly opposite directions cancel each other out.
Thus, C_{x} = 0. This implies that C = C_{y}. To solve for C_{y}, we need to solve for A_{y} and B_{y}
Note that A_{y} and B_{y} point in the same direction. Convention dictates we define up as positive, and down negative.
Thus,Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 13 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
13. Question
13. An ant is crawling up a set of stairs. Each stair measures 2 meters tall by 1 meter deep. There are 4 stairs. The ant crawls from the bottom to the top of the set of stairs. On the fourth stair, the ant climbs all the way to the end of the stair. What is the approximate ratio of the ant’s distance traveled to its displacement?Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Stair Measurements: 2 m tall x 1m deep
 Total # of Stairs: 4
Solving for: approximate ratio of ant’s distance traveled to its displacement
Step 2. Next, let’s work backwards. What do we need in order to solve for the approximate ratio of ant’s distance traveled to its displacement?
We need to calculate 1) the ant’s distance traveled and 2) the ant’s displacement.
Note: the distance traveled will always be ≥ displacement. So we can cross out any answer options that have displacement > distance traveled.
1) The ant’s distance traveled: each stair measures 2 m tall x 1m deep. Thus, for each stair the ant climbs, it has traveled (2 m + 1 m) = 3m. There are 4 stairs total. 3m x 4 = 12 m
2) The ant’s displacement: the ant’s displacement is the distance from the beginning of the bottom stair to the back of the final stair. We need to calculate the x and y components of the ant’s displacement, then use the Pythagorean equation to solve for the displacement (hypotenuse).
D_{x} = 4 stairs x 1 m deep / stair = 4m
D_{y} = 4 stairs x 2 m tall / stair = 8 mStep 3. Plug D_{x} and D_{y} into the Pythagorean equation to solve for the displacement (hypotenuse)
(D_{x})^{2} + (D_{y})^{2} = (D)^{2}
(4m)^{2} + (8m)^{2} = (D)^{2}
16m^{2} + 64m^{2} = (D)^{2}
80m^{2} = (D)^{2}
D = √80 m^{2} ≅ 9 mStep 4. Plug in the values we just calculated to determine the the approximate ratio of ant’s distance traveled to its displacement.
 Distance traveled: 12m
 Displacement ≅ 9 m
 Ratio distance traveled : displacement = 12m : 9 m = 4:3
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Stair Measurements: 2 m tall x 1m deep
 Total # of Stairs: 4
Solving for: approximate ratio of ant’s distance traveled to its displacement
Step 2. Next, let’s work backwards. What do we need in order to solve for the approximate ratio of ant’s distance traveled to its displacement?
We need to calculate 1) the ant’s distance traveled and 2) the ant’s displacement.
Note: the distance traveled will always be ≥ displacement. So we can cross out any answer options that have displacement > distance traveled.
1) The ant’s distance traveled: each stair measures 2 m tall x 1m deep. Thus, for each stair the ant climbs, it has traveled (2 m + 1 m) = 3m. There are 4 stairs total. 3m x 4 = 12 m
2) The ant’s displacement: the ant’s displacement is the distance from the beginning of the bottom stair to the back of the final stair. We need to calculate the x and y components of the ant’s displacement, then use the Pythagorean equation to solve for the displacement (hypotenuse).
D_{x} = 4 stairs x 1 m deep / stair = 4m
D_{y} = 4 stairs x 2 m tall / stair = 8 mStep 3. Plug D_{x} and D_{y} into the Pythagorean equation to solve for the displacement (hypotenuse)
(D_{x})^{2} + (D_{y})^{2} = (D)^{2}
(4m)^{2} + (8m)^{2} = (D)^{2}
16m^{2} + 64m^{2} = (D)^{2}
80m^{2} = (D)^{2}
D = √80 m^{2} ≅ 9 mStep 4. Plug in the values we just calculated to determine the the approximate ratio of ant’s distance traveled to its displacement.
 Distance traveled: 12m
 Displacement ≅ 9 m
 Ratio distance traveled : displacement = 12m : 9 m = 4:3

Question 14 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
14. Question
14. A ball is shot with velocity V = 10 √2 m/s at an angle 45^{o} from the horizontal. Determine the x and y components of the ball’s velocity.Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Velocity (V): 10 √2 m/s
 θ = 45^{o}
Solving for: the x and y components of the ball’s velocity
Step 2. Next, let’s work backwards. What do we need in order to solve for the x and y components of the ball’s velocity (V_{x} and V_{y})?
For many different types of problems on the OAT, you will need to know that:
1. V_{x} = V cos (θ)
2. V_{y} = V sin (θ)where “V” is any vector. These are based on trigonometry. If you’re not already, get very familiar with these equations.
Step 3. We know all the variables we need to solve for V_{x} and V_{y}. Plug them in to get your answer.
Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Velocity (V): 10 √2 m/s
 θ = 45^{o}
Solving for: the x and y components of the ball’s velocity
Step 2. Next, let’s work backwards. What do we need in order to solve for the x and y components of the ball’s velocity (V_{x} and V_{y})?
For many different types of problems on the OAT, you will need to know that:
1. V_{x} = V cos (θ)
2. V_{y} = V sin (θ)where “V” is any vector. These are based on trigonometry. If you’re not already, get very familiar with these equations.
Step 3. We know all the variables we need to solve for V_{x} and V_{y}. Plug them in to get your answer.
Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 15 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
15. Question
15. A man is launched out of a canon with velocity V = 100 m/s at an angle 60^{o} from the horizontal. Determine the x and y components of the the man’s velocity.Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Velocity (V): 100 m/s
 θ = 60^{o}
Solving for: the x and y components of the ball’s velocity
Step 2. Next, let’s work backwards. What do we need in order to solve for the x and y components of the man’s velocity (V_{x} and V_{y})?
For many different types of problems on the OAT, you will need to know that:
1. V_{x} = V cos (θ)
2. V_{y} = V sin (θ)where “V” is any vector. These are based on trigonometry. If you’re not already, get very familiar with these equations.
Step 3. We know all the variables we need to solve for V_{x} and V_{y}. Plug them in to get your answer.
Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Velocity (V): 100 m/s
 θ = 60^{o}
Solving for: the x and y components of the ball’s velocity
Step 2. Next, let’s work backwards. What do we need in order to solve for the x and y components of the man’s velocity (V_{x} and V_{y})?
For many different types of problems on the OAT, you will need to know that:
1. V_{x} = V cos (θ)
2. V_{y} = V sin (θ)where “V” is any vector. These are based on trigonometry. If you’re not already, get very familiar with these equations.
Step 3. We know all the variables we need to solve for V_{x} and V_{y}. Plug them in to get your answer.
Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 16 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
16. Question
16. Lewis and Clark are plotting their journey across the United States. Using their map, they plot the last 3 days of their journey, starting from base camp. Day 1: Start: (0,0) Stop: (4,4) Day 2: Start: (4,4) Stop: (12,7) Day 3: Start: (12,7) Stop: (12, 5) What is the magnitude of their displacement from base camp at the end of day 3?Correct / You marked this questionStep 1. First, let’s write down a quick sketch, and what we are solving for:
Solving for: the magnitude of their displacement from base camp at the end of day 3
Step 2. Let’s work backwards. How do we solve for the magnitude of their displacement from base camp at the end of day 3?
If you look at our sketch, you can see that the displacement vector goes from (0,0) to (12,5). As you can see, the displacement vector is independent of the path taken to arrive there. Thus, you really only need the start point and end point to solve this problem.
With the Pythagorean equation, we can use the x and y components of the displacement vector to solve for the magnitude of the displacement vector. The x component (D_{x}) of the displacement vector will go from (0,0) to (12,0), and the y component (D_{y}) will go from (12,0) to (12,5).
Step 3. Let’s calculate D_{x} and D_{y} to be able to plug in later.
 D_{x} = 12 – 0 = 12
 D_{y} = 0 – 5 = 5
When in doubt, there’s no shame in counting the little boxes 🙂
Remember, we are only solving for the magnitude of the displacement vector, so you can ignore the direction of D_{x} and D_{y} in this case.
Step 3. Plug D_{x} and D_{y} into the Pythagorean equation to solve for D.
Pythagorean Equation: x^{2} + y^{2} = z^{2}
(D_{x})^{2} + (D_{y})^{2} = (D)^{2}
(12)^{2} + (5)^{2} = (D)^{2}
144 + 25 = (D)^{2}
169 = (D)^{2}
D = √169 = 13Incorrect / You marked this questionStep 1. First, let’s write down a quick sketch, and what we are solving for:
Solving for: the magnitude of their displacement from base camp at the end of day 3
Step 2. Let’s work backwards. How do we solve for the magnitude of their displacement from base camp at the end of day 3?
If you look at our sketch, you can see that the displacement vector goes from (0,0) to (12,5). As you can see, the displacement vector is independent of the path taken to arrive there. Thus, you really only need the start point and end point to solve this problem.
With the Pythagorean equation, we can use the x and y components of the displacement vector to solve for the magnitude of the displacement vector. The x component (D_{x}) of the displacement vector will go from (0,0) to (12,0), and the y component (D_{y}) will go from (12,0) to (12,5).
Step 3. Let’s calculate D_{x} and D_{y} to be able to plug in later.
 D_{x} = 12 – 0 = 12
 D_{y} = 0 – 5 = 5
When in doubt, there’s no shame in counting the little boxes 🙂
Remember, we are only solving for the magnitude of the displacement vector, so you can ignore the direction of D_{x} and D_{y} in this case.
Step 3. Plug D_{x} and D_{y} into the Pythagorean equation to solve for D.
Pythagorean Equation: x^{2} + y^{2} = z^{2}
(D_{x})^{2} + (D_{y})^{2} = (D)^{2}
(12)^{2} + (5)^{2} = (D)^{2}
144 + 25 = (D)^{2}
169 = (D)^{2}
D = √169 = 13 
Question 17 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
17. Question
17. Three vectors of magnitudes A =15 N, B = 15 N, and C = 20N are applied to a system. If each of the vectors can be aligned to point any horizontal or vertical direction, what is the minimum magnitude of the resultant vector?Correct / You marked this questionStep 1. First, let’s write down a quick sketch, and what we are solving for:
Solving for: the minimum magnitude of the resultant vector
Step 2. This one is a bit tricky. The best way I know how to solve these problems is through trial and error.
Scenario 1: A and B point directly opposite, cancelling each other out.
The net force of A+ B = 0 N. Thus, our resultant vector, no matter what direction you point it in, will = C and have a magnitude of 20N.
Scenario 2: A and C point directly opposite. B points perpendicular.
The net force of A+ C = 5 N. Per the Pythagorean equation, our resultant vector will be
You are not expected to be able to solve this without a calculator, but you can deduce that the answer will be greater than 15.
Scenario 3: A and B point in the same direction. C points in the opposite direction.
The magnitude of the resultant vector = (15 N + 15 N ) – 20 N = 10 N . This is the smallest resultant vector magnitude we can obtain.
Strategically: even if you thought there might be a smaller magnitude you could obtain, you have tried three primary scenarios. You have put in enough time into this question. Mark it for review and move on.
Incorrect / You marked this questionStep 1. First, let’s write down a quick sketch, and what we are solving for:
Solving for: the minimum magnitude of the resultant vector
Step 2. This one is a bit tricky. The best way I know how to solve these problems is through trial and error.
Scenario 1: A and B point directly opposite, cancelling each other out.
The net force of A+ B = 0 N. Thus, our resultant vector, no matter what direction you point it in, will = C and have a magnitude of 20N.
Scenario 2: A and C point directly opposite. B points perpendicular.
The net force of A+ C = 5 N. Per the Pythagorean equation, our resultant vector will be
You are not expected to be able to solve this without a calculator, but you can deduce that the answer will be greater than 15.
Scenario 3: A and B point in the same direction. C points in the opposite direction.
The magnitude of the resultant vector = (15 N + 15 N ) – 20 N = 10 N . This is the smallest resultant vector magnitude we can obtain.
Strategically: even if you thought there might be a smaller magnitude you could obtain, you have tried three primary scenarios. You have put in enough time into this question. Mark it for review and move on.

Question 18 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
18. Question
18. Michael Phelps needs to cross a river. The river is flowing 15 m/s East. Michael Phelps jumps in and swims at an angle 60^{o} South of West. How fast does Michael Phelps have to swim in order to move ONLY across the river, and not upstream or downstream?Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Velocity of River (R) = 15 m/s East
 θ = 60^{o}
Solving for: Michael Phelps’ velocity (V_{MP}) in order to move ONLY across the river, and not upstream or downstream
Step 2. Next, let’s work backwards. What does it mean for Michael Phelps to move ONLY across the river, and not upstream or downstream?
If Michael Phelps is not moving upstream or downstream, this implies that the horizontal component of his velocity is equal and opposite to the velocity of the river, and thus net horizontal velocity = 0 m/s . In mathematical terms, that would mean:
R = V_{MP} cos (θ)
Step 3. Let’s plug in given values to solve for Michael Phelps’ Velocity (V_{MP})
15 m/s = V_{MP} cos (60)
15 m/s = V_{MP} (1/2)
V_{MP} = 30 m/sIncorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Velocity of River (R) = 15 m/s East
 θ = 60^{o}
Solving for: Michael Phelps’ velocity (V_{MP}) in order to move ONLY across the river, and not upstream or downstream
Step 2. Next, let’s work backwards. What does it mean for Michael Phelps to move ONLY across the river, and not upstream or downstream?
If Michael Phelps is not moving upstream or downstream, this implies that the horizontal component of his velocity is equal and opposite to the velocity of the river, and thus net horizontal velocity = 0 m/s . In mathematical terms, that would mean:
R = V_{MP} cos (θ)
Step 3. Let’s plug in given values to solve for Michael Phelps’ Velocity (V_{MP})
15 m/s = V_{MP} cos (60)
15 m/s = V_{MP} (1/2)
V_{MP} = 30 m/s 
Question 19 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
19. Question
19. A girl throws a 5 kg ball 45^{o} from the horizontal. The ball travels 50 m horizontally. Which of the following best depicts the direction of the ball’s acceleration?Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass = 5 kg
 θ = 45^{o}
 Horizontal Distance traveled: 50m
Solving for: the direction of the ball’s acceleration
Step 2. Next, let’s work backwards. What do we need in order to solve for the direction of the ball’s acceleration?
For problems like this, it is critical to understand that there is NO acceleration in the x (horizontal) direction (a_{x} = 0).
Thus, we can rule out any answer options which having a horizontal component (i.e. are not pointing straight up or straight down).
From here, we can deduce from intuition that, due to gravity, the ball’s acceleration will be pointing downwards (↓)
In this problem, there was a lot of extraneous information provided. The “working backwards” method will help you avoid wasting your time on this.
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass = 5 kg
 θ = 45^{o}
 Horizontal Distance traveled: 50m
Solving for: the direction of the ball’s acceleration
Step 2. Next, let’s work backwards. What do we need in order to solve for the direction of the ball’s acceleration?
For problems like this, it is critical to understand that there is NO acceleration in the x (horizontal) direction (a_{x} = 0).
Thus, we can rule out any answer options which having a horizontal component (i.e. are not pointing straight up or straight down).
From here, we can deduce from intuition that, due to gravity, the ball’s acceleration will be pointing downwards (↓)
In this problem, there was a lot of extraneous information provided. The “working backwards” method will help you avoid wasting your time on this.

Question 20 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
20. Question
20. A superhero must stop a runaway train from falling off the edge of a cliff. The train starts at velocity 12 m/s. The superhero causes the train to decelerate at 2 m/s^{2}. There are 40 meters before the train falls off the cliff. How many meters before the edge of the cliff does the superhero stop the train?Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Initial Velocity Train (v_{i}) = 12 m/s
 Acceleration = – 2 m/s^{2}
 Final Velocity (v_{f}) = 0 m/s
 Distance to Edge of Cliff = 40 m
Solving for: meters before the edge of the cliff train stops (D_{s})
Step 2. Next, let’s work backwards. What do we need in order to solve for meters before the edge of the cliff train stops?
In order to solve for this, we need to know how many meters the train traveled with the super hero working to stop it.
There are two kinematics equations you might think to use here:
1. displacement = v_{i}t + ^{1}/_{2}at^{2}
2. v_{f}^{2} = v_{i}^{2} + 2adDetermining which one to use is a matter of seeing which variables you are already given, or can solve for.
 In the first equation, we are given v_{i} and a. That means we have two unknown variables (displacement and time). We could use the following equation to solve for time: v_{f} = v_{i} + at
 In second equation, we are given all variables (v_{f}, v_{i}, a) except d (displacement) which is what we are solving for.
Thus, you can use either equation #1 or #2 to obtain the correct answer, but #2 is more direct. Given the tight time constraints, #2 is preferable.
Step 3. Plug in our known variables into equation #2 to determine displacement of train.
v_{f}^{2} = v_{i}^{2} + 2ad
0 = (12 m/s)^{2} + 2(2 m/s^{2})(d)
0 = 144 m^{2}/s^{2} – (4m/s^{2}) d
d = (144 m^{2}/s^{2}) / (4m/s^{2}) = 36 mStep 4. Remember that we’re solving for meters before edge of cliff that train stops. Thus:
Distance to edge of cliff – displacement of train = distance before edge of cliff that train stops.
40 m – 36 m = 4 mIncorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Initial Velocity Train (v_{i}) = 12 m/s
 Acceleration = – 2 m/s^{2}
 Final Velocity (v_{f}) = 0 m/s
 Distance to Edge of Cliff = 40 m
Solving for: meters before the edge of the cliff train stops (D_{s})
Step 2. Next, let’s work backwards. What do we need in order to solve for meters before the edge of the cliff train stops?
In order to solve for this, we need to know how many meters the train traveled with the super hero working to stop it.
There are two kinematics equations you might think to use here:
1. displacement = v_{i}t + ^{1}/_{2}at^{2}
2. v_{f}^{2} = v_{i}^{2} + 2adDetermining which one to use is a matter of seeing which variables you are already given, or can solve for.
 In the first equation, we are given v_{i} and a. That means we have two unknown variables (displacement and time). We could use the following equation to solve for time: v_{f} = v_{i} + at
 In second equation, we are given all variables (v_{f}, v_{i}, a) except d (displacement) which is what we are solving for.
Thus, you can use either equation #1 or #2 to obtain the correct answer, but #2 is more direct. Given the tight time constraints, #2 is preferable.
Step 3. Plug in our known variables into equation #2 to determine displacement of train.
v_{f}^{2} = v_{i}^{2} + 2ad
0 = (12 m/s)^{2} + 2(2 m/s^{2})(d)
0 = 144 m^{2}/s^{2} – (4m/s^{2}) d
d = (144 m^{2}/s^{2}) / (4m/s^{2}) = 36 mStep 4. Remember that we’re solving for meters before edge of cliff that train stops. Thus:
Distance to edge of cliff – displacement of train = distance before edge of cliff that train stops.
40 m – 36 m = 4 m 
Question 21 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
21. Question
21. An 80 kg Olympic diver trips and falls off of a tall diving board. It takes him 7 seconds to hit the water. How tall is the diving board? Assume gravitational acceleration = 10 m/s^{2}Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Initial Velocity Diver (v_{i}) = 0 m/s
 Acceleration = 10 m/s^{2}
 Time: 7 seconds
 Mass: 80 kg
Solving for: height of diving board
Step 2. Next, let’s work backwards. What do we need to solve for the height of the diving board?
Note: objects in free fall do so independently of their mass (assuming we neglect air resistance, which is the case for problems on the OAT) . Thus, you can disregard the diver’s mass from the get go.
If it takes the diver 7 seconds to hit the water, then the height of the diving board will be the distance travelled in these 7 seconds. Let’s refer to one of our handydandy kinematics equations:
 d = v_{i}t + ^{1}/_{2}at^{2}
Plug in our given values to the above kinematics equations to solve for d (height of diving board).
d = v_{i}t + ^{1}/_{2}at^{2}
d = (0 m/s) (7s) + ^{1}/_{2}(10 m/s^{2}) (7s)^{2}
d = (5 m/s^{2}) (49 s^{2})
d = 245 mIncorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Initial Velocity Diver (v_{i}) = 0 m/s
 Acceleration = 10 m/s^{2}
 Time: 7 seconds
 Mass: 80 kg
Solving for: height of diving board
Step 2. Next, let’s work backwards. What do we need to solve for the height of the diving board?
Note: objects in free fall do so independently of their mass (assuming we neglect air resistance, which is the case for problems on the OAT) . Thus, you can disregard the diver’s mass from the get go.
If it takes the diver 7 seconds to hit the water, then the height of the diving board will be the distance travelled in these 7 seconds. Let’s refer to one of our handydandy kinematics equations:
 d = v_{i}t + ^{1}/_{2}at^{2}
Plug in our given values to the above kinematics equations to solve for d (height of diving board).
d = v_{i}t + ^{1}/_{2}at^{2}
d = (0 m/s) (7s) + ^{1}/_{2}(10 m/s^{2}) (7s)^{2}
d = (5 m/s^{2}) (49 s^{2})
d = 245 m 
Question 22 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
22. Question
22. Three preoptometry students try out an experiment. At the exact same time, they each throw a bowling ball upwards from the bottom of three different cliffs: Student A throws a 100 kg bowling ball 105 m/s directly upward to a 100 m cliff.
 Student B throws a 80 kg bowling ball 40 m/s directly upward to a 60 m cliff.
 Student C throws a 70 kg bowling ball 20 m/s directly upward to a 20 m cliff.
Correct / You marked this questionStep 1. First, let’s write down our variables and what we are solving for:
 Mass Ball A (M_{A}) = 100 kg
 Initial Velocity Ball A (V_{iA}) = 105 m/s
 Height Cliff A = 100 m
 Mass Ball B (M_{B}) = 80 kg
 Initial Velocity Ball B (V_{iB}) = 40 m/s
 Height Cliff B = 60 m
 Mass Ball C (M_{C}) = 70 kg
 Initial Velocity Ball C (V_{iC}) = 20 m/s
 Height Cliff C = 20 m
Solving for: the order the bowling balls reaches the top, from first to last
Step 2. Next, let’s work backwards. What do we need to solve for the order the bowling balls reach the cliff, from first to last?
Note: objects in free fall do so independently of their mass (assuming we neglect air resistance, which is the case for problems on the OAT) . Thus, you can disregard the mass of the bowling balls from the get go.
The order the bowling balls reach the top of the respective cliffs will depend on the time it takes each to reach their cliffs.
To solve for the time it takes each ball to reach their cliffs, let’s refer to one of our handydandy kinematics equations:
 d = v_{i}t + ^{1}/_{2}at^{2}
Step 3. Plug in our given values to the above kinematics equations to solve for t for each bowling ball:
Ball A:
d = v_{i}t + ^{1}/_{2}at^{2}
100 m= (105 m/s) (t) – ^{1}/_{2}(10 m/s^{2}) (t^{2})
100 m= (105 m/s) (t) – (5 m/s^{2}) (t^{2})
t = 1 secondBall B:
d = v_{i}t + ^{1}/_{2}at^{2}
60 m= (40 m/s) (t) – ^{1}/_{2}(10 m/s^{2}) (t^{2})
60 m= (40 m/s) (t) – (5 m/s^{2}) (t^{2})
t = 2 secondsBall C:
d = v_{i}t + ^{1}/_{2}at^{2}
20 m= (20 m/s) (t) – ^{1}/_{2}(10 m/s^{2}) (t^{2})
20 m= (20 m/s) (t) – (5 m/s^{2}) (t^{2})
t = 2 secondsIncorrect / You marked this questionStep 1. First, let’s write down our variables and what we are solving for:
 Mass Ball A (M_{A}) = 100 kg
 Initial Velocity Ball A (V_{iA}) = 105 m/s
 Height Cliff A = 100 m
 Mass Ball B (M_{B}) = 80 kg
 Initial Velocity Ball B (V_{iB}) = 40 m/s
 Height Cliff B = 60 m
 Mass Ball C (M_{C}) = 70 kg
 Initial Velocity Ball C (V_{iC}) = 20 m/s
 Height Cliff C = 20 m
Solving for: the order the bowling balls reaches the top, from first to last
Step 2. Next, let’s work backwards. What do we need to solve for the order the bowling balls reach the cliff, from first to last?
Note: objects in free fall do so independently of their mass (assuming we neglect air resistance, which is the case for problems on the OAT) . Thus, you can disregard the mass of the bowling balls from the get go.
The order the bowling balls reach the top of the respective cliffs will depend on the time it takes each to reach their cliffs.
To solve for the time it takes each ball to reach their cliffs, let’s refer to one of our handydandy kinematics equations:
 d = v_{i}t + ^{1}/_{2}at^{2}
Step 3. Plug in our given values to the above kinematics equations to solve for t for each bowling ball:
Ball A:
d = v_{i}t + ^{1}/_{2}at^{2}
100 m= (105 m/s) (t) – ^{1}/_{2}(10 m/s^{2}) (t^{2})
100 m= (105 m/s) (t) – (5 m/s^{2}) (t^{2})
t = 1 secondBall B:
d = v_{i}t + ^{1}/_{2}at^{2}
60 m= (40 m/s) (t) – ^{1}/_{2}(10 m/s^{2}) (t^{2})
60 m= (40 m/s) (t) – (5 m/s^{2}) (t^{2})
t = 2 secondsBall C:
d = v_{i}t + ^{1}/_{2}at^{2}
20 m= (20 m/s) (t) – ^{1}/_{2}(10 m/s^{2}) (t^{2})
20 m= (20 m/s) (t) – (5 m/s^{2}) (t^{2})
t = 2 seconds 
Question 23 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
23. Question
23. From ground level, a man shoots a gun directly upwards with an initial velocity of 100 m/s. How much time does it take for the bullet to ascend to maximum height, and fall back to the ground? Assume gravitational acceleration = 10 m/s^{2}Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Initial Velocity (v_{i}) = 100 m/s
 Acceleration = 10 m/s^{2} (downward)
Solving for: time it takes bullet to ascend and descend back to ground
Step 2. Next, let’s work backwards. What do we need to solve for the time it takes bullet to ascend and descend back to ground?
We’re definitely going to need to use one (or more) of our kinematics equations to solve for this.
Let’s took a look at our options:
We don’t know how far the bullet travels (d), so #1 has two unknown variables (d and t). We can rule that one out.
#2 and #3 both require that we known v_{f}. You know v_{f} at the top of the bullet’s journey is 0 m/s. Let’s call this velocity at the top of the bullet’s journey v_{f1} and the velocity when it returns to the ground v_{f2}.
We can deduce that, because the force of gravity is constant, and we are neglecting other forces such as air resistance, the bullet will lose velocity on the way up at the same rate that it gains velocity on the way down. Thus, v_{i} = v_{f}
In this case, #2 is the most direct route to solve our problem, as #3 requires that we determine displacement.
Step 3. Break the bullet’s journey in to ascent and descent. Plug our given variables into equation #2 to determine time in flight.
Ascent:
v_{f1} = v_{i1} + at
0 m/s = 100 m/s + (10 m/s^{2}) t
t = 10 secondsDescent:
v_{f2} = v_{i2} + at
100 m/s = 0 m/s – (10 m/s^{2}) t
t = 10 secondsTotal flight time: 10 s + 10 s = 20 seconds
Step 4. There are a number of important takeaways from this problem:
 You can often use any or multiple of the kinematics equations to eventually get to the correct answer, but usually there is one clear quickest route. As time is of the essence, make sure this is the route you take!
 An object shot directly upwards will return to the ground with the same initial velocity as when first fired. Also, the time it takes to ascend will equal the time it takes to descend.
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Initial Velocity (v_{i}) = 100 m/s
 Acceleration = 10 m/s^{2} (downward)
Solving for: time it takes bullet to ascend and descend back to ground
Step 2. Next, let’s work backwards. What do we need to solve for the time it takes bullet to ascend and descend back to ground?
We’re definitely going to need to use one (or more) of our kinematics equations to solve for this.
Let’s took a look at our options:
We don’t know how far the bullet travels (d), so #1 has two unknown variables (d and t). We can rule that one out.
#2 and #3 both require that we known v_{f}. You know v_{f} at the top of the bullet’s journey is 0 m/s. Let’s call this velocity at the top of the bullet’s journey v_{f1} and the velocity when it returns to the ground v_{f2}.
We can deduce that, because the force of gravity is constant, and we are neglecting other forces such as air resistance, the bullet will lose velocity on the way up at the same rate that it gains velocity on the way down. Thus, v_{i} = v_{f}
In this case, #2 is the most direct route to solve our problem, as #3 requires that we determine displacement.
Step 3. Break the bullet’s journey in to ascent and descent. Plug our given variables into equation #2 to determine time in flight.
Ascent:
v_{f1} = v_{i1} + at
0 m/s = 100 m/s + (10 m/s^{2}) t
t = 10 secondsDescent:
v_{f2} = v_{i2} + at
100 m/s = 0 m/s – (10 m/s^{2}) t
t = 10 secondsTotal flight time: 10 s + 10 s = 20 seconds
Step 4. There are a number of important takeaways from this problem:
 You can often use any or multiple of the kinematics equations to eventually get to the correct answer, but usually there is one clear quickest route. As time is of the essence, make sure this is the route you take!
 An object shot directly upwards will return to the ground with the same initial velocity as when first fired. Also, the time it takes to ascend will equal the time it takes to descend.

Question 24 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
24. Question
24. Based on the below graph, determine the velocity from 30 seconds to 40 seconds.
Correct / You marked this questionStep 1. We are provided a sketch. Let’s write down what we’re solving for:
Solving for: velocity from 30 seconds to 40 seconds
Step 2. Next, let’s work backwards. What do we need to solve for the velocity from 30 seconds to 40 seconds?
We are given a position versus time graph. It is important that you know how to interpret this type of graph:
 The slope over a certain time interval = average velocity
 Nonlinear slope over a certain time interval time graph implies nonzero acceleration over this time interval
Thus, to solve for the (average) velocity from 30s to 40s, we simply need to determine the slope in that time frame.
Step 3. Determine slope between 30s to 40s.
Slope = rise / run.
Rise = 0m – 50m = – 50m
Run = 40s – 30s = 10sSlope = Velocity = 50m/10s = 5m/s
It is important to take into account if the slope (and thus velocity) is negative or positive.
Incorrect / You marked this questionStep 1. We are provided a sketch. Let’s write down what we’re solving for:
Solving for: velocity from 30 seconds to 40 seconds
Step 2. Next, let’s work backwards. What do we need to solve for the velocity from 30 seconds to 40 seconds?
We are given a position versus time graph. It is important that you know how to interpret this type of graph:
 The slope over a certain time interval = average velocity
 Nonlinear slope over a certain time interval time graph implies nonzero acceleration over this time interval
Thus, to solve for the (average) velocity from 30s to 40s, we simply need to determine the slope in that time frame.
Step 3. Determine slope between 30s to 40s.
Slope = rise / run.
Rise = 0m – 50m = – 50m
Run = 40s – 30s = 10sSlope = Velocity = 50m/10s = 5m/s
It is important to take into account if the slope (and thus velocity) is negative or positive.

Question 25 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
25. Question
25. Based on the below graph, determine the average velocity from 0 to 40 seconds.
Correct / You marked this questionStep 1. We are provided a sketch. Let’s write down what we’re solving for:
Solving for: velocity from 0 to 40 seconds
Step 2. Next, let’s work backwards. What do we need to solve for the average velocity from 0 to 40 seconds?
We are given a position versus time graph. It is important that you know how to interpret this type of graph:
 The slope over a certain time interval = average velocity
 Nonlinear slope over a certain time interval time graph implies nonzero acceleration over this time interval
Thus, to solve for the average velocity from 0s to 40s, we simply need to determine the slope in that time frame.
You may be confused by the fact that the position points from 0 to 40s are not connected. However, as we are determining average velocity, this does not change the way we proceed with the calculation. All we need to known is the start point and end point!
Step 3. Determine slope between 0s to 40s.
Slope = rise / run.
Rise = 40m – 60m = 20m
Run = 40s – 0s = 40 secondsSlope = Velocity = 20m/40s = –^{1}/_{2} m/s
It is important to take into account if the slope (and thus velocity) is negative or positive.
Incorrect / You marked this questionStep 1. We are provided a sketch. Let’s write down what we’re solving for:
Solving for: velocity from 0 to 40 seconds
Step 2. Next, let’s work backwards. What do we need to solve for the average velocity from 0 to 40 seconds?
We are given a position versus time graph. It is important that you know how to interpret this type of graph:
 The slope over a certain time interval = average velocity
 Nonlinear slope over a certain time interval time graph implies nonzero acceleration over this time interval
Thus, to solve for the average velocity from 0s to 40s, we simply need to determine the slope in that time frame.
You may be confused by the fact that the position points from 0 to 40s are not connected. However, as we are determining average velocity, this does not change the way we proceed with the calculation. All we need to known is the start point and end point!
Step 3. Determine slope between 0s to 40s.
Slope = rise / run.
Rise = 40m – 60m = 20m
Run = 40s – 0s = 40 secondsSlope = Velocity = 20m/40s = –^{1}/_{2} m/s
It is important to take into account if the slope (and thus velocity) is negative or positive.

Question 26 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
26. Question
26. A boy takes a slingshot and shoots a 3 kg rock horizontally off of his of a cliff of 25 meters in height, at an initial velocity of 10 m/s. How far horizontally from the cliff will the rock land? (Use g = 10 m/s^{2})Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass = 3 kg
 Initial Horizontal Velocity (v_{ix}) = 10 m/s
 Initial Vertical Velocity (v_{iy}) = 0 m/s
 Height Cliff (d_{y}) = 25 meters
 Gravitational Acceleration = 10 m/s^{2}
Solving for: Horizontal distance of rock from cliff (d_{x})
Step 2. Next, let’s work backwards. What do we need in order to solve for horizontal distance from cliff (d_{x})?
For problems like this, it is critical to understand two things:
1. There is NO acceleration in the x (horizontal) direction (a_{x} = 0)
2. The time it takes for the object to fall from the top of cliff to the ground, due to gravitational acceleration (i.e. t in the y direction) will be the same time the object travels horizontally (i.e. t in the x direction).We know that d_{x} = v_{ix}t + ^{1}/_{2}a_{x}t^{2}.
If a_{x} = 0
Then, d_{x} = v_{ix}t.We already are given v_{ix} = 10m/s, so to solve for d_{x}, we need to determine t (the time the rock is in flight)
Step 3. Next, we need to determine how to solve for t (the time the rock is in flight)
Note: The mass of the rock (3 kg) is irrelevant in this problem! The OAT will have you assume ideal conditions for objects in free fall, so you can neglect mass/ air resistance.
We know that d_{y} = v_{iy}t + ^{1}/_{2}a_{y}t^{2}.
a_{y}= 10 m/s^{2}
v_{iy}= 0 m/s
d = 25 metersThen, 25m = ^{1}/_{2}(10) (t^{2})
t^{2} = 5
t = √5Step 4. Plug t back into our equation for d_{x}
d_{x} = v_{ix}t
d_{x} = (10 m/s) (√5 s)
d_{x} = 10 √5 mIncorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass = 3 kg
 Initial Horizontal Velocity (v_{ix}) = 10 m/s
 Initial Vertical Velocity (v_{iy}) = 0 m/s
 Height Cliff (d_{y}) = 25 meters
 Gravitational Acceleration = 10 m/s^{2}
Solving for: Horizontal distance of rock from cliff (d_{x})
Step 2. Next, let’s work backwards. What do we need in order to solve for horizontal distance from cliff (d_{x})?
For problems like this, it is critical to understand two things:
1. There is NO acceleration in the x (horizontal) direction (a_{x} = 0)
2. The time it takes for the object to fall from the top of cliff to the ground, due to gravitational acceleration (i.e. t in the y direction) will be the same time the object travels horizontally (i.e. t in the x direction).We know that d_{x} = v_{ix}t + ^{1}/_{2}a_{x}t^{2}.
If a_{x} = 0
Then, d_{x} = v_{ix}t.We already are given v_{ix} = 10m/s, so to solve for d_{x}, we need to determine t (the time the rock is in flight)
Step 3. Next, we need to determine how to solve for t (the time the rock is in flight)
Note: The mass of the rock (3 kg) is irrelevant in this problem! The OAT will have you assume ideal conditions for objects in free fall, so you can neglect mass/ air resistance.
We know that d_{y} = v_{iy}t + ^{1}/_{2}a_{y}t^{2}.
a_{y}= 10 m/s^{2}
v_{iy}= 0 m/s
d = 25 metersThen, 25m = ^{1}/_{2}(10) (t^{2})
t^{2} = 5
t = √5Step 4. Plug t back into our equation for d_{x}
d_{x} = v_{ix}t
d_{x} = (10 m/s) (√5 s)
d_{x} = 10 √5 m 
Question 27 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
27. Question
27. Ricky Bobby, moving at 360 km/hr in his 700 kg race car, slams on his brakes to narrowly avoid crashing into his friend Cal, who had walked onto the track. It takes him 10 seconds to stop. What is the average acceleration of Ricky’s car?Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass = 700 kg
 Initial Velocity (v_{i}) = 360 km/hr
 Final Velocity (v_{f}) = 0 m/s
 Stopping time (t) = 10 s
Solving for: Average acceleration of the car
Step 2. Next, let’s work backwards. What do we need in to solve for average acceleration of the car?
Average acceleration: a_{avg} = (v_{f} – v_{i}) / t
You may be tempted to go to our beloved F = m*a equation, but this is not the right scenario to use it! We are not given any information about the force required to bring Ricky’s car to rest. Thus, we would have two unknown variables in the F = m*a equation. The mass of the car was included in the problem to throw you off.
Plugging in our given variables, we get:
a_{avg} = (0 – 360 km/hr) / 10 seconds
Pause. The units look odd. From our answer choices, we want acceleration in units of m/s^{2}. To make sure our answer yields these units, convert 360 km/hr into m/s. Take your time and get it right.
1 hr = 60 seconds * 60 = 3600 seconds
360 km = 1000 * 360 m = 360,000 m
360 km/hr = 360,000 m / 3600 s = 100 m/s
a_{avg} = (0 – 100 m/s) / 10 seconds
a_{avg} = – 10 m/s^{2}
Step 3. Last, let’s do a quick sanity check. Does it make sense that we got a negative answer?
Ricky’s car is going from really fast, to a stop. I.e. his car is decelerating. So yes, it does make sense that our acceleration value is negative. If average acceleration was positive, that would imply that his car was speeding up.
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass = 700 kg
 Initial Velocity (v_{i}) = 360 km/hr
 Final Velocity (v_{f}) = 0 m/s
 Stopping time (t) = 10 s
Solving for: Average acceleration of the car
Step 2. Next, let’s work backwards. What do we need in to solve for average acceleration of the car?
Average acceleration: a_{avg} = (v_{f} – v_{i}) / t
You may be tempted to go to our beloved F = m*a equation, but this is not the right scenario to use it! We are not given any information about the force required to bring Ricky’s car to rest. Thus, we would have two unknown variables in the F = m*a equation. The mass of the car was included in the problem to throw you off.
Plugging in our given variables, we get:
a_{avg} = (0 – 360 km/hr) / 10 seconds
Pause. The units look odd. From our answer choices, we want acceleration in units of m/s^{2}. To make sure our answer yields these units, convert 360 km/hr into m/s. Take your time and get it right.
1 hr = 60 seconds * 60 = 3600 seconds
360 km = 1000 * 360 m = 360,000 m
360 km/hr = 360,000 m / 3600 s = 100 m/s
a_{avg} = (0 – 100 m/s) / 10 seconds
a_{avg} = – 10 m/s^{2}
Step 3. Last, let’s do a quick sanity check. Does it make sense that we got a negative answer?
Ricky’s car is going from really fast, to a stop. I.e. his car is decelerating. So yes, it does make sense that our acceleration value is negative. If average acceleration was positive, that would imply that his car was speeding up.

Question 28 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
28. Question
28. A car company is testing how quickly it’s new model car will reach 25 m/s (approximately 60 mph). The car starts from a stand still and accelerates at 5 m/s^{2}. How many seconds will it take for the car to reach 25 m/s?Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Initial Velocity (v_{i}) = 0 m/s
 Acceleration = 5 m/s^{2}
 Final Velocity (v_{f}) = 25 m/s
Solving for: time (in seconds) for car to reach 25 m/s
Step 2. Next, let’s work backwards. What do we need to solve for the time (in seconds) it takes for the car to reach 25 m/s?
We are given v_{f}, v_{i}, and a, and asked to solve for t. The following kinematics equation will work splendidly:
v_{f} = v_{i} + at
Next, plug in our given variables to solve for t.
v_{f} = v_{i} + at
25m/s = 0 m/s + (5m/s^{2}) t
25 m/s = (5 m/s^{2}) t
t = 5 secondsThere are a couple of ways you could get this calculation WRONG. Please AVOID the following:
 You see m/s^{2} in 5 m/s^{2} and square the 5, to yield 25 m/s^{2} (or 25 m^{2}/s^{4} if you’re really off). This would yield t = 1 second
 You calculate at as at^{2}. This would yield t = √5 seconds.
Two rules of thumb to use to avoid these mistakes are:
 When in doubt, always check your units. In this case: a*t should have corresponding units: (m/s^{2}) (s) = m/s.
 We use t^{2} in our kinematics equation for displacement, but NOT for the final velocity equation. Relevant kinematics equations are listed below for reference.
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Initial Velocity (v_{i}) = 0 m/s
 Acceleration = 5 m/s^{2}
 Final Velocity (v_{f}) = 25 m/s
Solving for: time (in seconds) for car to reach 25 m/s
Step 2. Next, let’s work backwards. What do we need to solve for the time (in seconds) it takes for the car to reach 25 m/s?
We are given v_{f}, v_{i}, and a, and asked to solve for t. The following kinematics equation will work splendidly:
v_{f} = v_{i} + at
Next, plug in our given variables to solve for t.
v_{f} = v_{i} + at
25m/s = 0 m/s + (5m/s^{2}) t
25 m/s = (5 m/s^{2}) t
t = 5 secondsThere are a couple of ways you could get this calculation WRONG. Please AVOID the following:
 You see m/s^{2} in 5 m/s^{2} and square the 5, to yield 25 m/s^{2} (or 25 m^{2}/s^{4} if you’re really off). This would yield t = 1 second
 You calculate at as at^{2}. This would yield t = √5 seconds.
Two rules of thumb to use to avoid these mistakes are:
 When in doubt, always check your units. In this case: a*t should have corresponding units: (m/s^{2}) (s) = m/s.
 We use t^{2} in our kinematics equation for displacement, but NOT for the final velocity equation. Relevant kinematics equations are listed below for reference.

Question 29 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
29. Question
29. An engineer is determining how powerful he needs to design a train’s brakes. The design specifications indicate that the train needs to be able to go from full speed (100 m/s) to a complete stop in 5 s. What is the deceleration that the train’s brakes must be able to cause?Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Initial Velocity (v_{i}) = 100 m/s
 Final Velocity (v_{f}) = 0 m/s
 Time (t) = 5 seconds
Solving for: deceleration required to stop train
Step 2. Next, let’s work backwards. What do we need to solve for the deceleration required to stop train?
We are given v_{f}, v_{i}, and t, and asked to solve for a (deceleration just means negative acceleration). The following kinematics equation will work like a charm:
v_{f} = v_{i} + at
Step 3. Next, plug in our given variables to solve for t.
v_{f} = v_{i} + at
0 m/s = 100 m/s + (a) (5 seconds)
100 m/s = (a) (5 seconds)
a = 20 m/s^{2} → deceleration = 20 m/s^{2}There are a couple of ways you could get this calculation WRONG. Please AVOID the following:
 You calculate a*t as a^{2}t. This would yield deceleration = 20 m/s^{2}
 You calculate a*t as at^{2}. This would yield deceleration = 4 m/s^{2}
Two rules of thumb to use to avoid these mistakes are:
 When in doubt, always check your units. (m/s)/(s) = m/s^{2}, which is the desired unit for acceleration here
 We use t^{2} in our kinematics equation for displacement, but NOT for the final velocity equation. Relevant kinematics equations are listed below for reference.
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Initial Velocity (v_{i}) = 100 m/s
 Final Velocity (v_{f}) = 0 m/s
 Time (t) = 5 seconds
Solving for: deceleration required to stop train
Step 2. Next, let’s work backwards. What do we need to solve for the deceleration required to stop train?
We are given v_{f}, v_{i}, and t, and asked to solve for a (deceleration just means negative acceleration). The following kinematics equation will work like a charm:
v_{f} = v_{i} + at
Step 3. Next, plug in our given variables to solve for t.
v_{f} = v_{i} + at
0 m/s = 100 m/s + (a) (5 seconds)
100 m/s = (a) (5 seconds)
a = 20 m/s^{2} → deceleration = 20 m/s^{2}There are a couple of ways you could get this calculation WRONG. Please AVOID the following:
 You calculate a*t as a^{2}t. This would yield deceleration = 20 m/s^{2}
 You calculate a*t as at^{2}. This would yield deceleration = 4 m/s^{2}
Two rules of thumb to use to avoid these mistakes are:
 When in doubt, always check your units. (m/s)/(s) = m/s^{2}, which is the desired unit for acceleration here
 We use t^{2} in our kinematics equation for displacement, but NOT for the final velocity equation. Relevant kinematics equations are listed below for reference.

Question 30 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
30. Question
30. A car accelerates from a complete stop to 200 m/s in 20 seconds, at an average acceleration of 10 m/s^{2}. t(0) = 0 m/s ; t(5) = 20 m/s ; t(10) = 60 m/s ; t(20) = 200 m/s Which graph best depicts the car’s acceleration as a function of time?Correct / You marked this questionStep 1. Let’s analyze this problem.
 a_{avg} = 10 m/s^{2}
 v_{i} = 0 m/s
 v_{f} = 200 m/s
 t(0) = 0 m/s ; t(5) = 20 m/s ; t(10) = 60 m/s ; t(20) = 200 m/s
Solving for: which graph best depicts acceleration of car
Step 2. Let’s work backwards. How can we determine which graph best depicts the acceleration of the car?
First, make sure you look carefully at the graphs’ axes. The y axis is acceleration, NOT velocity. Thus, we are actually plotting the acceleration over time, NOT looking at the slope of a velocity versus time graph.
We are given that the average acceleration over 20 s is 10 m/s^{2}. However this doesn’t tell us much about how the overall graph would look.
Alternatively, we are given the car’s velocity at t= 0, 5, 10, and 20. We can use this information to determine the car’s average acceleration over these specific intervals, and then see which graph lines up most closely.
Step 3. Calculate the car’s average acceleration from 0 to 5s, 5s to 10s, and 10s to 20s
a_{avg} = (v_{f} – v_{i}) / t
a_{avg} (05s) = (20 m/s – 0 m/s) / 5s = 4 m/s^{2}
a_{avg} (510s) = (60 m/s – 20 m/s) / 5s = 8 m/s^{2}
a_{avg} (1020s) = (200 m/s – 60 m/s) / 10s = 14 m/s^{2}Step 4. Analyze your findings from the above calculations
 We can deduce that the car does not have constant acceleration, otherwise a_{avg} (05s) would equal a_{avg} (510s) would equal a_{avg} (1020s). This rules out graph A.
 We can deduce (probably from the get go) that the car is not decelerating. This rules out graph B.
 We can deduce that the car’s acceleration is not increasing linearly, otherwise a_{avg} (05s) would equal X * a_{avg} (510s) and a_{avg} (510s) would equal X * a_{avg} (1020s). This rules out graph C.
 We can deduce that the car’s acceleration is increasing at an exponential rate. Graph D jumps towards its max acceleration relatively quickly, and then slowly approaches the maximum. Thus, we can rule out Graph D.
Based on our deductions and process of elimination, Graph E is the correct answer.
Incorrect / You marked this questionStep 1. Let’s analyze this problem.
 a_{avg} = 10 m/s^{2}
 v_{i} = 0 m/s
 v_{f} = 200 m/s
 t(0) = 0 m/s ; t(5) = 20 m/s ; t(10) = 60 m/s ; t(20) = 200 m/s
Solving for: which graph best depicts acceleration of car
Step 2. Let’s work backwards. How can we determine which graph best depicts the acceleration of the car?
First, make sure you look carefully at the graphs’ axes. The y axis is acceleration, NOT velocity. Thus, we are actually plotting the acceleration over time, NOT looking at the slope of a velocity versus time graph.
We are given that the average acceleration over 20 s is 10 m/s^{2}. However this doesn’t tell us much about how the overall graph would look.
Alternatively, we are given the car’s velocity at t= 0, 5, 10, and 20. We can use this information to determine the car’s average acceleration over these specific intervals, and then see which graph lines up most closely.
Step 3. Calculate the car’s average acceleration from 0 to 5s, 5s to 10s, and 10s to 20s
a_{avg} = (v_{f} – v_{i}) / t
a_{avg} (05s) = (20 m/s – 0 m/s) / 5s = 4 m/s^{2}
a_{avg} (510s) = (60 m/s – 20 m/s) / 5s = 8 m/s^{2}
a_{avg} (1020s) = (200 m/s – 60 m/s) / 10s = 14 m/s^{2}Step 4. Analyze your findings from the above calculations
 We can deduce that the car does not have constant acceleration, otherwise a_{avg} (05s) would equal a_{avg} (510s) would equal a_{avg} (1020s). This rules out graph A.
 We can deduce (probably from the get go) that the car is not decelerating. This rules out graph B.
 We can deduce that the car’s acceleration is not increasing linearly, otherwise a_{avg} (05s) would equal X * a_{avg} (510s) and a_{avg} (510s) would equal X * a_{avg} (1020s). This rules out graph C.
 We can deduce that the car’s acceleration is increasing at an exponential rate. Graph D jumps towards its max acceleration relatively quickly, and then slowly approaches the maximum. Thus, we can rule out Graph D.
Based on our deductions and process of elimination, Graph E is the correct answer.

Question 31 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
31. Question
31. You step on a scale on Earth, and it reads 50 kg. Planet X is ^{1}/_{4} the mass of Earth, and ^{1}/_{4} the radius of Earth. What is the ratio of your weight on Planet X to your weight on Earth?Correct / You marked this questionStep 1. First, let’s write down our variables and what we are solving for:
 Weight Earth = 50 kg
 Mass Planet X = ^{1}/_{4} Earth
 Radius Planet X = ^{1}/_{4} Earth
Solving for: ratio of your weight on Planet X to your weight on Earth
Step 2. Next, let’s work backwards. What do we need to solve for the ratio of your weight on Plant X to your weight on Earth?
This is a perfect situation to utilize our gravitational force between two bodies equation.
If we solve for the gravitational force on Earth, then the gravitational force on Planet X, we can compare the ratio of the two.
Step 3. Plug in our known values to the gravitational force equation to determine the gravitational force for each planet.
Note: your mass on Earth will be exactly the same on Earth and Planet X. Mass does not change based on location!
F(Earth) = (G (your mass) (mass Earth)) / (radius of earth)^{2}
F(Planet X) = (G (your mass) (^{1}/_{4} mass Earth)) / (^{1}/_{4} radius of earth)^{2}
Step 4. Compare the ratio of the two gravitational forces.
F(Planet X) will be (^{1}/_{4} value of Earth – due to mass) / (^{1}/_{16} value of Earth – due to radius)
(^{1}/_{4}) / (^{1}/_{16}) = 4. This implies the ratio of your weight on Planet X to Earth will be 4:1
Incorrect / You marked this questionStep 1. First, let’s write down our variables and what we are solving for:
 Weight Earth = 50 kg
 Mass Planet X = ^{1}/_{4} Earth
 Radius Planet X = ^{1}/_{4} Earth
Solving for: ratio of your weight on Planet X to your weight on Earth
Step 2. Next, let’s work backwards. What do we need to solve for the ratio of your weight on Plant X to your weight on Earth?
This is a perfect situation to utilize our gravitational force between two bodies equation.
If we solve for the gravitational force on Earth, then the gravitational force on Planet X, we can compare the ratio of the two.
Step 3. Plug in our known values to the gravitational force equation to determine the gravitational force for each planet.
Note: your mass on Earth will be exactly the same on Earth and Planet X. Mass does not change based on location!
F(Earth) = (G (your mass) (mass Earth)) / (radius of earth)^{2}
F(Planet X) = (G (your mass) (^{1}/_{4} mass Earth)) / (^{1}/_{4} radius of earth)^{2}
Step 4. Compare the ratio of the two gravitational forces.
F(Planet X) will be (^{1}/_{4} value of Earth – due to mass) / (^{1}/_{16} value of Earth – due to radius)
(^{1}/_{4}) / (^{1}/_{16}) = 4. This implies the ratio of your weight on Planet X to Earth will be 4:1

Question 32 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
32. Question
32. You step on a scale on Earth, and it reads 40 kg. Planet X has a mass 10 times greater than Earth, and a radius 2 times greater than Earth. What is the ratio of your mass on Planet X to your mass on Earth?Correct / You marked this questionStep 1: Before we write down our variables and what we are solving for, let’s make sure we read the question carefully.
“What is the ratio of your mass on Planet X to your mass on Earth?”
We quickly remember that an object’s mass is a constant value, regardless of location!
Thus, the ratio of mass on Planet X to mass on Earth is 1:1.
Incorrect / You marked this questionStep 1: Before we write down our variables and what we are solving for, let’s make sure we read the question carefully.
“What is the ratio of your mass on Planet X to your mass on Earth?”
We quickly remember that an object’s mass is a constant value, regardless of location!
Thus, the ratio of mass on Planet X to mass on Earth is 1:1.

Question 33 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
33. Question
33. A car decelerates from 15 m/s to 5 m/s in 5 seconds. How far did the car travel in this time period?Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
Initial Velocity (v_{i}) = 15 m/s
Final Velocity (v_{f}) = 5 m/s
Time (t) = 5 secondsSolving for: distance traveled by car in this time period
Step 2. Next, let’s work backwards. What do we need to solve for the distance traveled by the car in this time period?
In general, we know the equation for displacement (which will be the same as distance traveled in this scenario) is:
We are given v_{i} and t. Thus, if we only had a (average acceleration) we could have our answer!
Step 3. Work backwards: how do we solve for a (average acceleration)?
Per our equations sheet, we know that:
Solving for a_{avg}:
a_{avg} = (v_{f} – v_{i}) / t
= (5 m/s – 15 m/s) / 5 s
= – 2 m/s^{2}Step 4. Plug a_{avg} back into our displacement equation to solve for distance traveled:
Solving for d:
d = v_{i}t + ^{1}/_{2}at^{2}
= 15 m/s (5 seconds) + ^{1}/_{2}(2 m/s^{2}) (5 seconds^{2})
= 75 m – 25 m
= 50 mIncorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
Initial Velocity (v_{i}) = 15 m/s
Final Velocity (v_{f}) = 5 m/s
Time (t) = 5 secondsSolving for: distance traveled by car in this time period
Step 2. Next, let’s work backwards. What do we need to solve for the distance traveled by the car in this time period?
In general, we know the equation for displacement (which will be the same as distance traveled in this scenario) is:
We are given v_{i} and t. Thus, if we only had a (average acceleration) we could have our answer!
Step 3. Work backwards: how do we solve for a (average acceleration)?
Per our equations sheet, we know that:
Solving for a_{avg}:
a_{avg} = (v_{f} – v_{i}) / t
= (5 m/s – 15 m/s) / 5 s
= – 2 m/s^{2}Step 4. Plug a_{avg} back into our displacement equation to solve for distance traveled:
Solving for d:
d = v_{i}t + ^{1}/_{2}at^{2}
= 15 m/s (5 seconds) + ^{1}/_{2}(2 m/s^{2}) (5 seconds^{2})
= 75 m – 25 m
= 50 m 
Question 34 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
34. Question
34. A car decelerates from at a uniform rate from 15 m/s to 5 m/s in 5 seconds. How far did the car travel in the 2^{nd} second of deceleration?Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Initial Velocity (v_{i}) = 15 m/s
 Final Velocity (v_{f}) = 5 m/s
 Time (t) = 2^{nd} second (t = 1)
Solving for: distance traveled by car in in the 2^{nd} second
Step 2. Next, let’s work backwards. What do we need to solve for the distance traveled by the car in the 2^{nd} second?
In general, we know the equation for displacement (which will be the same as distance traveled in this scenario) is:
We are given v_{i} and t. Thus, we need average acceleration (i.e. deceleration in this scenario) to be able to use our displacement equation.
Step 3. Work backwards: how do we solve for a (average acceleration)?
Per our equations sheet, we know that:
Solving for a_{avg}:
a_{avg} = (v_{f} – v_{i}) / t
= (5 m/s – 15 m/s) / 5 s
= – 2 m/s^{2}Step 4. We can now use our displacement equation. But we must ask ourselves: what is the question is asking for? The distance traveled by car in the 2^{nd} second
For problems like this, I like to write down the data at 1 second intervals:
Step 5. Next, let’s use our displacement equation to solve for distance traveled by car in the 2^{nd} second
The first second is from t = 0 to t = 1. The 2nd second is from t = 1 to t = 2.
Plug a_{avg} back into our displacement equation to solve for distance traveled:
Solving for d:
d = v_{i}t + ^{1}/_{2}at^{2}
= 13 m/s (1 second) + ^{1}/_{2}(2 m/s^{2}) (1 second)^{2}
= 13 m – 1 m
= 12 mIncorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Initial Velocity (v_{i}) = 15 m/s
 Final Velocity (v_{f}) = 5 m/s
 Time (t) = 2^{nd} second (t = 1)
Solving for: distance traveled by car in in the 2^{nd} second
Step 2. Next, let’s work backwards. What do we need to solve for the distance traveled by the car in the 2^{nd} second?
In general, we know the equation for displacement (which will be the same as distance traveled in this scenario) is:
We are given v_{i} and t. Thus, we need average acceleration (i.e. deceleration in this scenario) to be able to use our displacement equation.
Step 3. Work backwards: how do we solve for a (average acceleration)?
Per our equations sheet, we know that:
Solving for a_{avg}:
a_{avg} = (v_{f} – v_{i}) / t
= (5 m/s – 15 m/s) / 5 s
= – 2 m/s^{2}Step 4. We can now use our displacement equation. But we must ask ourselves: what is the question is asking for? The distance traveled by car in the 2^{nd} second
For problems like this, I like to write down the data at 1 second intervals:
Step 5. Next, let’s use our displacement equation to solve for distance traveled by car in the 2^{nd} second
The first second is from t = 0 to t = 1. The 2nd second is from t = 1 to t = 2.
Plug a_{avg} back into our displacement equation to solve for distance traveled:
Solving for d:
d = v_{i}t + ^{1}/_{2}at^{2}
= 13 m/s (1 second) + ^{1}/_{2}(2 m/s^{2}) (1 second)^{2}
= 13 m – 1 m
= 12 m 
Question 35 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
35. Question
35. What is Newton’s First Law of Motion?Correct / You marked this questionLet’s look at each answer option individually:
[A] For every action, there is an equal and opposite reaction.
This is Newton’s 3^{rd} Law of Motion
[B] Force is equal to the change in momentum per change in time. For a constant mass, force equals mass times acceleration.
This is Newton’s 2^{nd} Law of Motion
[C] Every object persists in its state of rest or uniform motion in a straight line unless it is compelled to change that state by forces impressed upon it.
This is the correct answer
[D] An object attracts other objects with a force directly proportional to the product of their masses, and inversely proportional to the square of the distance between their centers.
This is Newton’s First Law of Gravitation
[E] Energy can neither be created nor destroyed in an isolated system.
This is the First Law of Thermodynamics
Incorrect / You marked this questionLet’s look at each answer option individually:
[A] For every action, there is an equal and opposite reaction.
This is Newton’s 3^{rd} Law of Motion
[B] Force is equal to the change in momentum per change in time. For a constant mass, force equals mass times acceleration.
This is Newton’s 2^{nd} Law of Motion
[C] Every object persists in its state of rest or uniform motion in a straight line unless it is compelled to change that state by forces impressed upon it.
This is the correct answer
[D] An object attracts other objects with a force directly proportional to the product of their masses, and inversely proportional to the square of the distance between their centers.
This is Newton’s First Law of Gravitation
[E] Energy can neither be created nor destroyed in an isolated system.
This is the First Law of Thermodynamics

Question 36 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
36. Question
36. What is Newton’s Second Law of Motion?Correct / You marked this questionLet’s look at each answer option individually:
[A] For every action, there is an equal and opposite reaction.
This is Newton’s Third Law of Motion
[B] Force is equal to the change in momentum per change in time. For a constant mass, force equals mass times acceleration.
This is the correct answer
[C] Energy can neither be created nor destroyed in an isolated system.
This is the First Law of Thermodynamics
[D] Every object persists in its state of rest or uniform motion in a straight line unless it is compelled to change that state by forces impressed upon it.
This is Newton’s First Law of Motion
[E] An object attracts other objects with a force directly proportional to the product of their masses, and inversely proportional to the square of the distance between their centers.
This is Newton’s First Law of Gravitation
Incorrect / You marked this questionLet’s look at each answer option individually:
[A] For every action, there is an equal and opposite reaction.
This is Newton’s Third Law of Motion
[B] Force is equal to the change in momentum per change in time. For a constant mass, force equals mass times acceleration.
This is the correct answer
[C] Energy can neither be created nor destroyed in an isolated system.
This is the First Law of Thermodynamics
[D] Every object persists in its state of rest or uniform motion in a straight line unless it is compelled to change that state by forces impressed upon it.
This is Newton’s First Law of Motion
[E] An object attracts other objects with a force directly proportional to the product of their masses, and inversely proportional to the square of the distance between their centers.
This is Newton’s First Law of Gravitation

Question 37 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
37. Question
37. What is Newton’s Third Law of Motion?Correct / You marked this questionLet’s look at each answer option individually:
[A] An object attracts other objects with a force directly proportional to the product of their masses, and inversely proportional to the square of the distance between their centers.
This is Newton’s First Law of Gravitation
[B] Force is equal to the change in momentum per change in time. For a constant mass, force equals mass times acceleration.
This is Newton’s Second Law of Motion
[C] Energy can neither be created nor destroyed in an isolated system.
This is the First Law of Thermodynamics
[D] Every object persists in its state of rest or uniform motion in a straight line unless it is compelled to change that state by forces impressed upon it.
This is Newton’s First Law of Motion
[E] For every action, there is an equal and opposite reaction.
This is the correct answer
Incorrect / You marked this questionLet’s look at each answer option individually:
[A] An object attracts other objects with a force directly proportional to the product of their masses, and inversely proportional to the square of the distance between their centers.
This is Newton’s First Law of Gravitation
[B] Force is equal to the change in momentum per change in time. For a constant mass, force equals mass times acceleration.
This is Newton’s Second Law of Motion
[C] Energy can neither be created nor destroyed in an isolated system.
This is the First Law of Thermodynamics
[D] Every object persists in its state of rest or uniform motion in a straight line unless it is compelled to change that state by forces impressed upon it.
This is Newton’s First Law of Motion
[E] For every action, there is an equal and opposite reaction.
This is the correct answer

Question 38 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
38. Question
38. Which of the following best exemplifies Newton’s First Law of Motion?Correct / You marked this questionLet’s look at each answer option individually:
[A] When a man pushes on a box with force F, the box exerts a force on the man of force F.
This exemplifies Newton’s 3^{rd} Law of Motion
[B] The force of a 1000 kg car accelerating at 10 m/s^{2} is 10,000 N.
This exemplifies Newton’s 2^{nd} Law of Motion
[C] When a soccer player kicks a ball down the field, it will eventually come to a stop due air resistance and the friction of the ground.
This is the correct answer
[D] The gravitational force an object experiences will be exponentially greater the farther away the object is from Earth.
This exemplifies Newton’s First Law of Gravitation
[E] If you drop a ball, the kinetic energy on its descent can never exceed the potential energy at its apex.
This exemplifies the First Law of Thermodynamics
Incorrect / You marked this questionLet’s look at each answer option individually:
[A] When a man pushes on a box with force F, the box exerts a force on the man of force F.
This exemplifies Newton’s 3^{rd} Law of Motion
[B] The force of a 1000 kg car accelerating at 10 m/s^{2} is 10,000 N.
This exemplifies Newton’s 2^{nd} Law of Motion
[C] When a soccer player kicks a ball down the field, it will eventually come to a stop due air resistance and the friction of the ground.
This is the correct answer
[D] The gravitational force an object experiences will be exponentially greater the farther away the object is from Earth.
This exemplifies Newton’s First Law of Gravitation
[E] If you drop a ball, the kinetic energy on its descent can never exceed the potential energy at its apex.
This exemplifies the First Law of Thermodynamics

Question 39 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
39. Question
39. Which of the following best exemplifies Newton’s Second Law of Motion?Correct / You marked this questionLet’s look at each answer option individually:
[A] When a man pushes on a box with force F, the box exerts a force on the man of force F.
This exemplifies Newton’s 3^{rd} Law of Motion
[B] The force of a 1000 kg car accelerating at 10 m/s^{2} is 10,000 N.
This is the correct answer
[C] If you drop a ball, the kinetic energy on its descent can never exceed the potential energy at its apex.
This exemplifies the First Law of Thermodynamics
[D] When a soccer player kicks a ball down the field, it will eventually come to a stop due air resistance and the friction of the ground.
This exemplifies Newton’s First Law of Motion
[E] The gravitational force an object experiences will be exponentially greater the farther away the object is from Earth.
This exemplifies Newton’s First Law of Gravitation
Incorrect / You marked this questionLet’s look at each answer option individually:
[A] When a man pushes on a box with force F, the box exerts a force on the man of force F.
This exemplifies Newton’s 3^{rd} Law of Motion
[B] The force of a 1000 kg car accelerating at 10 m/s^{2} is 10,000 N.
This is the correct answer
[C] If you drop a ball, the kinetic energy on its descent can never exceed the potential energy at its apex.
This exemplifies the First Law of Thermodynamics
[D] When a soccer player kicks a ball down the field, it will eventually come to a stop due air resistance and the friction of the ground.
This exemplifies Newton’s First Law of Motion
[E] The gravitational force an object experiences will be exponentially greater the farther away the object is from Earth.
This exemplifies Newton’s First Law of Gravitation

Question 40 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
40. Question
40. Which of the following best exemplifies Newton’s Third Law of Motion?Correct / You marked this questionLet’s look at each answer option individually:
[A] The gravitational force an object experiences will be exponentially greater the farther away the object is from Earth.
This exemplifies Newton’s First Law of Gravitation
[B] The force of a 1000 kg car accelerating at 10 m/s^{2} is 10,000 N.
This exemplifies Newton’s 2^{nd} Law of Motion
[C] If you drop a ball, the kinetic energy on its descent can never exceed the potential energy at its apex.
This exemplifies the First Law of Thermodynamics
[D] When a soccer player kicks a ball down the field, it will eventually come to a stop due air resistance and the friction of the ground.
This exemplifies Newton’s First Law of Motion
[E] When a man pushes on a box with force F, the box exerts a force on the man of force F.
This is the correct answer
Incorrect / You marked this questionLet’s look at each answer option individually:
[A] The gravitational force an object experiences will be exponentially greater the farther away the object is from Earth.
This exemplifies Newton’s First Law of Gravitation
[B] The force of a 1000 kg car accelerating at 10 m/s^{2} is 10,000 N.
This exemplifies Newton’s 2^{nd} Law of Motion
[C] If you drop a ball, the kinetic energy on its descent can never exceed the potential energy at its apex.
This exemplifies the First Law of Thermodynamics
[D] When a soccer player kicks a ball down the field, it will eventually come to a stop due air resistance and the friction of the ground.
This exemplifies Newton’s First Law of Motion
[E] When a man pushes on a box with force F, the box exerts a force on the man of force F.
This is the correct answer

Question 41 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
41. Question
41. A superhero must stop a runaway train. The train has a mass of 1000 kg and starts at 5 m/s. The superhero has 51 m to stop the train, and stops it with 1 m to spare. What is the magnitude of the force he uses to stop the train?Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Initial Velocity Train (v_{i}) = 5 m/s
 Final Velocity (v_{f}) = 0 m/s
 Distance Traveled by Train = 51m1m = 50 m
Solving for: What is the magnitude of the force used by the superhero to stop the train
Step 2. Next, let’s work backwards. What do we need to be able to solve for the magnitude of the force used by the superhero to stop the train?
We know that Force = Mass * Acceleration.
We are given the mass of the train. So if we solve for acceleration (deceleration in this case), we can determine the force used to stop the train.
Step 3. Next, let’s work backwards. How can we solve for the acceleration (i.e. deceleration) of the train?
Look at the variables we are given, and then compare that against our equation sheet. We are given v_{f}, v_{i}, displacement / distance traveled, and we are solving for a.
Step 4. Solve for a, then plug into F = m X a
v_{f}^{2} = v_{i}^{2} + 2ad
0 m^{2}/s^{2} = (5 m/s)^{2} + 2(a)(50m)
0 m^{2}/s^{2} = 25 m^{2}/s^{2} + (100m) a
a = 0.25 m/s^{2}F = m * a
F = (1000 kg) * (0.25 m/s^{2})
F = 250NIncorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Initial Velocity Train (v_{i}) = 5 m/s
 Final Velocity (v_{f}) = 0 m/s
 Distance Traveled by Train = 51m1m = 50 m
Solving for: What is the magnitude of the force used by the superhero to stop the train
Step 2. Next, let’s work backwards. What do we need to be able to solve for the magnitude of the force used by the superhero to stop the train?
We know that Force = Mass * Acceleration.
We are given the mass of the train. So if we solve for acceleration (deceleration in this case), we can determine the force used to stop the train.
Step 3. Next, let’s work backwards. How can we solve for the acceleration (i.e. deceleration) of the train?
Look at the variables we are given, and then compare that against our equation sheet. We are given v_{f}, v_{i}, displacement / distance traveled, and we are solving for a.
Step 4. Solve for a, then plug into F = m X a
v_{f}^{2} = v_{i}^{2} + 2ad
0 m^{2}/s^{2} = (5 m/s)^{2} + 2(a)(50m)
0 m^{2}/s^{2} = 25 m^{2}/s^{2} + (100m) a
a = 0.25 m/s^{2}F = m * a
F = (1000 kg) * (0.25 m/s^{2})
F = 250N 
Question 42 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
42. Question
42. If the net force on an object is tripled and the mass remains the same, what is the effect on the object’s acceleration?Correct / You marked this questionBased on our net force equation:
F = m * a
3F = m(3a)If we triple the force (and hold the mass constant) this will cause the acceleration to triple as well.
Incorrect / You marked this questionBased on our net force equation:
F = m * a
3F = m(3a)If we triple the force (and hold the mass constant) this will cause the acceleration to triple as well.

Question 43 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
43. Question
43. If a car accelerates from 0 to v_{f} with acceleration a and displacement d, what would the car’s relative displacement be if it acceleration from 0 to 2v_{f} with the same acceleration a?Correct / You marked this questionStep 1. Let’s start by identifying which equation this question is referencing and what the question is actually asking for:
If we start from 0, double the velocity, and hold acceleration constant, what will be the effect on displacement (d)?
Step 2. Let’s compare the two cases:
Case 1:
v_{f}^{2} = v_{i}^{2} + 2ad
v_{f}^{2} = 2ad
d= v_{f}^{2} / 2aCase 2:
(2v_{f})^{2} = v_{i}^{2} + 2ad
4v_{f}^{2} = 2ad
d = 4v_{f}^{2} / 2aAs you can see, d has quadrupled from case 1 to case 2. It can be helpful to plug in actual values (use whatever #’s you’d like, as long as you hold them constant from Case 1 to Case 2). Use the simplest #’s that will work.
Incorrect / You marked this questionStep 1. Let’s start by identifying which equation this question is referencing and what the question is actually asking for:
If we start from 0, double the velocity, and hold acceleration constant, what will be the effect on displacement (d)?
Step 2. Let’s compare the two cases:
Case 1:
v_{f}^{2} = v_{i}^{2} + 2ad
v_{f}^{2} = 2ad
d= v_{f}^{2} / 2aCase 2:
(2v_{f})^{2} = v_{i}^{2} + 2ad
4v_{f}^{2} = 2ad
d = 4v_{f}^{2} / 2aAs you can see, d has quadrupled from case 1 to case 2. It can be helpful to plug in actual values (use whatever #’s you’d like, as long as you hold them constant from Case 1 to Case 2). Use the simplest #’s that will work.

Question 44 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
44. Question
44. A car is driving 50 m/s and slams on the brakes, which apply a stopping force F. If the same car is driving 100 m/s, how much farther would it take to the car to stop? Let d = original stopping distanceCorrect / You marked this questionStep 1. Let’s start by identifying which equation this question is referencing and what the question is actually asking for:
If we decelerate at double the starting velocity, how much longer will stopping distance be?
Step 2. Let’s compare the two cases:
Case 1:
v_{f}^{2} = v_{i}^{2} + 2ad
0 = v_{i}^{2} + 2ad
d= v_{i}^{2} / 2aCase 2:
v_{f}^{2} = (2v_{i})^{2} + 2ad
0 = 4v_{i}^{2} + 2ad
d = 4v_{i}^{2} / 2aAs you can see, d has quadrupled from case 1 to case 2. It can be helpful to plug in actual values (use whatever #’s you’d like, as long as you hold them constant from Case 1 to Case 2). Use the simplest #’s that will work.
HOWEVER, let’s be careful we note what the problem is asking for: how much farther it would take to the car to stop. If our original stopping distance is d, and our displacement has QUADRUPLE, then it takes 3d farther to stop.
Incorrect / You marked this questionStep 1. Let’s start by identifying which equation this question is referencing and what the question is actually asking for:
If we decelerate at double the starting velocity, how much longer will stopping distance be?
Step 2. Let’s compare the two cases:
Case 1:
v_{f}^{2} = v_{i}^{2} + 2ad
0 = v_{i}^{2} + 2ad
d= v_{i}^{2} / 2aCase 2:
v_{f}^{2} = (2v_{i})^{2} + 2ad
0 = 4v_{i}^{2} + 2ad
d = 4v_{i}^{2} / 2aAs you can see, d has quadrupled from case 1 to case 2. It can be helpful to plug in actual values (use whatever #’s you’d like, as long as you hold them constant from Case 1 to Case 2). Use the simplest #’s that will work.
HOWEVER, let’s be careful we note what the problem is asking for: how much farther it would take to the car to stop. If our original stopping distance is d, and our displacement has QUADRUPLE, then it takes 3d farther to stop.

Question 45 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
45. Question
45. A 80 kg man pushes a 10 kg box with a force of 20 N. What is the force the box exerts on the man?Correct / You marked this questionThis is a conceptual question. Newton’s Third Law says that for every action, there is an equal and opposite reaction. This implies that if the man pushes on the box with a force of 20 N, the box pushes back on the man with an equal force of 20 N.
Incorrect / You marked this questionThis is a conceptual question. Newton’s Third Law says that for every action, there is an equal and opposite reaction. This implies that if the man pushes on the box with a force of 20 N, the box pushes back on the man with an equal force of 20 N.

Question 46 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
46. Question
46. A plane is free falling out of control. A superhero flies in to save the day. He pulls directly upward on the plane. Under which condition does the plane reach terminal velocity?Correct / You marked this questionThis is conceptual question. Terminal velocity is maximum velocity attained by an object as it falls. An object will continue to accelerate downward until an upward force matches the downward force of gravity. So terminal velocity occurs when an object in free fall reaches acceleration = 0.
 When F_{gravity} > F_{superhero} the plane is still continuing to accelerate downward. It has not reached maximum velocity.
 When F_{gravity} < F_{superhero} the object would have stopped falling and started to rise. The plane stops in mid air, and thus never experiences this condition.
 When F_{gravity} = F_{superhero} there is 0 net force on the plane, and thus the plane’s acceleration= 0. The plane has reached terminal velocity.
Incorrect / You marked this questionThis is conceptual question. Terminal velocity is maximum velocity attained by an object as it falls. An object will continue to accelerate downward until an upward force matches the downward force of gravity. So terminal velocity occurs when an object in free fall reaches acceleration = 0.
 When F_{gravity} > F_{superhero} the plane is still continuing to accelerate downward. It has not reached maximum velocity.
 When F_{gravity} < F_{superhero} the object would have stopped falling and started to rise. The plane stops in mid air, and thus never experiences this condition.
 When F_{gravity} = F_{superhero} there is 0 net force on the plane, and thus the plane’s acceleration= 0. The plane has reached terminal velocity.

Question 47 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
47. Question
47. A boy drops a water balloon from his roof (46.5 m high) onto his little brother’s head. His brother is 1.5 m tall. Assume gravitational acceleration = 10 m/s^{2} How long does it take for the balloon to drop from the roof to splash on his brother’s head?Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Initial Velocity Water Balloon (v_{i}) = 0 m/s
 Acceleration = 10 m/s^{2}
 Distance Dropped = 46.5 m – 1.5 m = 45 m
Solving for: time it takes for balloon to fall onto brother’s head
Step 2. Next, let’s work backwards. What do we need to solve for the time it takes for balloon to fall onto brother’s head?
Let’s refer to one of our handydandy kinematics equations:
d = v_{i}t + ^{1}/_{2}at^{2}
Step 3. Plug in our given values to the above kinematics equations to solve for t
d = v_{i}t + ^{1}/_{2}at^{2}
45m = (0 m/s) (t) + ^{1}/_{2}(10 m/s^{2}) (t^{2})
45m = (5 m/s^{2}) (t^{2})
9s^{2} = t^{2}
t = 3 sIncorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Initial Velocity Water Balloon (v_{i}) = 0 m/s
 Acceleration = 10 m/s^{2}
 Distance Dropped = 46.5 m – 1.5 m = 45 m
Solving for: time it takes for balloon to fall onto brother’s head
Step 2. Next, let’s work backwards. What do we need to solve for the time it takes for balloon to fall onto brother’s head?
Let’s refer to one of our handydandy kinematics equations:
d = v_{i}t + ^{1}/_{2}at^{2}
Step 3. Plug in our given values to the above kinematics equations to solve for t
d = v_{i}t + ^{1}/_{2}at^{2}
45m = (0 m/s) (t) + ^{1}/_{2}(10 m/s^{2}) (t^{2})
45m = (5 m/s^{2}) (t^{2})
9s^{2} = t^{2}
t = 3 s 
Question 48 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
48. Question
48. A car accelerates from 12 m/s to 30 m/s over 378 m. What is the car’s acceleration over this time period?Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Initial Velocity (v_{i}) = 12 m/s
 Final Velocity (v_{f}) = 30 m/s
 Distance Traveled = 378 m
Solving for: car’s acceleration
Step 2. Next, let’s work backwards. What do we need to solve for the car’s acceleration?
We are given v_{f}, v_{i}, d, and asked to solve for a. The following kinematics equation will work like a charm:
v_{f}^{2} = v_{i}^{2} + 2ad
Step 3. Next, plug in our given variables to solve for a.
v_{f}^{2} = v_{i}^{2} + 2ad
(30 m/s)^{2} = (12 m/s)^{2} + 2a(378m)
900 m^{2}/s^{2} = (144 m^{2}/s^{2}) + (756m)a
756 m^{2}/s^{2} = (756m)a
a = 1 m/s^{2}Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Initial Velocity (v_{i}) = 12 m/s
 Final Velocity (v_{f}) = 30 m/s
 Distance Traveled = 378 m
Solving for: car’s acceleration
Step 2. Next, let’s work backwards. What do we need to solve for the car’s acceleration?
We are given v_{f}, v_{i}, d, and asked to solve for a. The following kinematics equation will work like a charm:
v_{f}^{2} = v_{i}^{2} + 2ad
Step 3. Next, plug in our given variables to solve for a.
v_{f}^{2} = v_{i}^{2} + 2ad
(30 m/s)^{2} = (12 m/s)^{2} + 2a(378m)
900 m^{2}/s^{2} = (144 m^{2}/s^{2}) + (756m)a
756 m^{2}/s^{2} = (756m)a
a = 1 m/s^{2} 
Question 49 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
49. Question
49. A 10 kg box is slides down a declined hill. The hill forms a 60^{o} angle with the horizontal. What is the normal force on the box? Assume gravitational acceleration = 10 m/s^{2}Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box = 10 kg
 θ = 60^{o}
 g = 10 m/s^{2}
Solving for: normal force (N)
Step 2. Next, let’s work backwards. What do we need to solve for the normal force (N) on the box?
Normal Force is the force that a surface exerts on an object that it is in contact with (in this case, the force that the hill exerts on the box). This force will be perpendicular to the surface of contact (in this case, the slant of the hill).
Thus, to solve for the Normal Force, we must solve for the perpendicular component of the object’s weight. In this case, we have termed this “W_{y}”Step 3. Next, let’s work backwards. How do we solve for W_{y}?
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2})
To find the y component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 30^{o}.
β_{1} = β_{2} = 30^{o}.
θ_{1} = θ_{2} = 60^{o}If you look at our sketch, you can see that the perpendicular component of weight (i.e W_{y}) will equal:
W_{y} = (mass)(g)(sinβ_{2})
W_{y} = (10kg)(10m/s^{2})(sin30)
W_{y} = N = 50 NKnowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box = 10 kg
 θ = 60^{o}
 g = 10 m/s^{2}
Solving for: normal force (N)
Step 2. Next, let’s work backwards. What do we need to solve for the normal force (N) on the box?
Normal Force is the force that a surface exerts on an object that it is in contact with (in this case, the force that the hill exerts on the box). This force will be perpendicular to the surface of contact (in this case, the slant of the hill).
Thus, to solve for the Normal Force, we must solve for the perpendicular component of the object’s weight. In this case, we have termed this “W_{y}”Step 3. Next, let’s work backwards. How do we solve for W_{y}?
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2})
To find the y component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 30^{o}.
β_{1} = β_{2} = 30^{o}.
θ_{1} = θ_{2} = 60^{o}If you look at our sketch, you can see that the perpendicular component of weight (i.e W_{y}) will equal:
W_{y} = (mass)(g)(sinβ_{2})
W_{y} = (10kg)(10m/s^{2})(sin30)
W_{y} = N = 50 NKnowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 50 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
50. Question
50. A 15 kg box is slides down a declined hill. The hill forms a 45^{o} angle with the horizontal. What is the normal force on the box? Assume gravitational acceleration = 10 m/s^{2}Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box = 15 kg
 θ = 45^{o}
 g = 10 m/s^{2}
Solving for = normal force (N)
Step 2. Next, let’s work backwards. What do we need to solve for the normal force (N) on the box?
Normal Force is the force that a surface exerts on an object that it is in contact with (in this case, the force that the hill exerts on the box). This force will be perpendicular to the surface of contact (in this case, the slant of the hill).Thus, to solve for the Normal Force, we must solve for the perpendicular component of the object’s weight. In this case, we have termed this “W_{y}”
Step 3. Next, let’s work backwards. How do we solve for W_{Y}?
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2})To find the y component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 45^{o}.
β_{1} = β_{2} = 45^{o}
θ_{1} = θ_{2} = 45^{o}If you look at our sketch, you can see that the perpendicular component of weight (i.e W_{Y}) will equal:
W_{y} = (mass)(g)(cosθ_{2})
W_{y} = (15kg)(10m/s^{2})(cos45)
W_{y} = N = ^{150}/_{√2} NKnowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box = 15 kg
 θ = 45^{o}
 g = 10 m/s^{2}
Solving for = normal force (N)
Step 2. Next, let’s work backwards. What do we need to solve for the normal force (N) on the box?
Normal Force is the force that a surface exerts on an object that it is in contact with (in this case, the force that the hill exerts on the box). This force will be perpendicular to the surface of contact (in this case, the slant of the hill).Thus, to solve for the Normal Force, we must solve for the perpendicular component of the object’s weight. In this case, we have termed this “W_{y}”
Step 3. Next, let’s work backwards. How do we solve for W_{Y}?
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2})To find the y component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 45^{o}.
β_{1} = β_{2} = 45^{o}
θ_{1} = θ_{2} = 45^{o}If you look at our sketch, you can see that the perpendicular component of weight (i.e W_{Y}) will equal:
W_{y} = (mass)(g)(cosθ_{2})
W_{y} = (15kg)(10m/s^{2})(cos45)
W_{y} = N = ^{150}/_{√2} NKnowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 51 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
51. Question
51. A 30 kg box is slides down a declined hill. The hill forms a 30^{o} angle with the horizontal. What is the normal force on the box? Assume gravitational acceleration = 10 m/s^{2}Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box = 30 kg
 θ = 30^{o}
 g = 10 m/s^{2}
Solving for: normal force (N)
Step 2. Next, let’s work backwards. What do we need to solve for the normal force (N) on the box?
Normal Force is the force that a surface exerts on an object that it is in contact with (in this case, the force that the hill exerts on the box). This force will be perpendicular to the surface of contact (in this case, the slant of the hill).
Thus, to solve for the Normal Force, we must solve for the perpendicular component of the object’s weight. In this case, we have termed this “W_{y}”Step 3. Next, let’s work backwards. How do we solve for W_{y}?
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2})
To find the y component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 60^{o}.
β_{1} = β_{1} = 60^{o}.
θ_{1} = θ_{1} = 30^{o}If you look at our sketch, you can see that the perpendicular component of weight (i.e W_{y}) will equal:
W_{y} = (mass)(g)(sinβ_{2})
W_{y} = (30kg)(10m/s^{2})(sin60)
W_{y} = N = 150 √3 NKnowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box = 30 kg
 θ = 30^{o}
 g = 10 m/s^{2}
Solving for: normal force (N)
Step 2. Next, let’s work backwards. What do we need to solve for the normal force (N) on the box?
Normal Force is the force that a surface exerts on an object that it is in contact with (in this case, the force that the hill exerts on the box). This force will be perpendicular to the surface of contact (in this case, the slant of the hill).
Thus, to solve for the Normal Force, we must solve for the perpendicular component of the object’s weight. In this case, we have termed this “W_{y}”Step 3. Next, let’s work backwards. How do we solve for W_{y}?
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2})
To find the y component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 60^{o}.
β_{1} = β_{1} = 60^{o}.
θ_{1} = θ_{1} = 30^{o}If you look at our sketch, you can see that the perpendicular component of weight (i.e W_{y}) will equal:
W_{y} = (mass)(g)(sinβ_{2})
W_{y} = (30kg)(10m/s^{2})(sin60)
W_{y} = N = 150 √3 NKnowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 52 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
52. Question
52. A 10 kg child slides down a frictionless declined slide. The slide forms a 60^{o} angle with the horizontal. Assuming the child started from rest, what is the child’s acceleration down the slide? Assume gravitational acceleration = 10 m/s^{2}Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Child= 10 kg
 θ = 60^{o}
 g = 10 m/s^{2}
Solving for: child’s acceleration down the slide
Step 2. Next, let’s work backwards. What do we need to solve for the child’s acceleration down the slide?
We need to use the F = m * a equation here.
We know that the net force in the y direction = 0 (i.e the normal force = the y component of the child’s weight). Thus, there is no acceleration in the y direction. So we are solving for a_{x}, where F_{x} = m * a_{x}
 Rearranging the above: a_{x} = F_{x} / m
 We know m (mass), so we need to determine F_{x}
Step 3. Next, let’s work backwards. How do we solve for F_{x}?
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2}). F_{x} is equivalent to W_{x} in our sketch.
To find the x component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 30^{o}.
β_{1} = β_{2} = 30^{o}.
θ_{1} = θ_{2} = 60^{o}If you look at our sketch, you can see that the parallel component of weight (i.e W_{x}) will equal:
W_{x} = (mass)(g)(cosβ_{2})
W_{x} = (10kg)(10m/s^{2})(cos30)
W_{x} = F_{x} = 50 √3 NPlug in F_{x} to solve for a_{x}:
a_{x}= F_{x} / m
a_{x}= (50 √3 N) / 10 kg
a_{x}= 5 √3 m/s^{2}Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Child= 10 kg
 θ = 60^{o}
 g = 10 m/s^{2}
Solving for: child’s acceleration down the slide
Step 2. Next, let’s work backwards. What do we need to solve for the child’s acceleration down the slide?
We need to use the F = m * a equation here.
We know that the net force in the y direction = 0 (i.e the normal force = the y component of the child’s weight). Thus, there is no acceleration in the y direction. So we are solving for a_{x}, where F_{x} = m * a_{x}
 Rearranging the above: a_{x} = F_{x} / m
 We know m (mass), so we need to determine F_{x}
Step 3. Next, let’s work backwards. How do we solve for F_{x}?
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2}). F_{x} is equivalent to W_{x} in our sketch.
To find the x component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 30^{o}.
β_{1} = β_{2} = 30^{o}.
θ_{1} = θ_{2} = 60^{o}If you look at our sketch, you can see that the parallel component of weight (i.e W_{x}) will equal:
W_{x} = (mass)(g)(cosβ_{2})
W_{x} = (10kg)(10m/s^{2})(cos30)
W_{x} = F_{x} = 50 √3 NPlug in F_{x} to solve for a_{x}:
a_{x}= F_{x} / m
a_{x}= (50 √3 N) / 10 kg
a_{x}= 5 √3 m/s^{2}Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 53 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
53. Question
53. A 15 kg child slides down a frictionless declined slide. The slide forms a 45^{o} angle with the horizontal. Assuming the child started from rest, what is the child’s acceleration down the slide? Assume gravitational acceleration = 10 m/s^{2}Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Child = 15 kg
 θ = 45^{o}
 g = 10 m/s^{2}
Solving for: child’s acceleration down the slide
Step 2. Next, let’s work backwards. What do we need to solve for the child’s acceleration down the slide?
We need to use the F = m * a equation here.
We know that the net force in the y direction = 0 (i.e the normal force = the y component of the child’s weight). Thus, there is no acceleration in the y direction. So we are solving for a_{x}, where F_{x} = m * a_{x}
 Rearranging the above: a_{x}= F_{x} / m
 We know m (mass), so we need to determine F_{x}
Step 3. Next, let’s work backwards. How do we solve for F_{x} (denoted W_{x} in our sketch)?
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2}). F_{x} is equivalent to a_{x} in our sketch.
To find the x component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 45^{o}.
β_{1} = β_{2} = 45^{o}.
θ_{1} = θ_{2} = 45^{o}If you look at our sketch, you can see that the parallel component of weight (i.e W_{x}) will equal:
W_{x} = (mass)(g)(cosβ_{2})
W_{x} = (15kg)(10m/s^{2})(cos 45)
W_{x} = F_{x} = ^{150}/_{√2} NStep 4. Plug in F_{x} to solve for a_{x}
a_{x}= F_{x} / m
a_{x}= (^{150}/_{√2} N) / 15 kg
a_{x}= ^{10}/_{√2} m/s^{2}Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Child = 15 kg
 θ = 45^{o}
 g = 10 m/s^{2}
Solving for: child’s acceleration down the slide
Step 2. Next, let’s work backwards. What do we need to solve for the child’s acceleration down the slide?
We need to use the F = m * a equation here.
We know that the net force in the y direction = 0 (i.e the normal force = the y component of the child’s weight). Thus, there is no acceleration in the y direction. So we are solving for a_{x}, where F_{x} = m * a_{x}
 Rearranging the above: a_{x}= F_{x} / m
 We know m (mass), so we need to determine F_{x}
Step 3. Next, let’s work backwards. How do we solve for F_{x} (denoted W_{x} in our sketch)?
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2}). F_{x} is equivalent to a_{x} in our sketch.
To find the x component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 45^{o}.
β_{1} = β_{2} = 45^{o}.
θ_{1} = θ_{2} = 45^{o}If you look at our sketch, you can see that the parallel component of weight (i.e W_{x}) will equal:
W_{x} = (mass)(g)(cosβ_{2})
W_{x} = (15kg)(10m/s^{2})(cos 45)
W_{x} = F_{x} = ^{150}/_{√2} NStep 4. Plug in F_{x} to solve for a_{x}
a_{x}= F_{x} / m
a_{x}= (^{150}/_{√2} N) / 15 kg
a_{x}= ^{10}/_{√2} m/s^{2}Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 54 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
54. Question
54. A 30 kg child slides down a frictionless declined slide. The slide forms a 30^{o} angle with the horizontal. Assuming the child started from rest, what is the child’s acceleration down the slide? Assume gravitational acceleration = 10 m/s^{2}Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Child = 30 kg
 θ = 30^{o}
 g = 10 m/s^{2}
Solving for: child’s acceleration down the slide
Step 2. Next, let’s work backwards. What do we need to solve for the child’s acceleration down the slide?
We need to use the F = m * a equation here.
We know that the net force in the y direction = 0 (i.e the normal force = the y component of the child’s weight). Thus, there there is no acceleration in the y direction. So we are solving for a_{x}, where F_{x} = m * a_{x}
 Rearranging the above: a_{x} = F_{x} / m
 We know m (mass), so we need to determine F_{x}
Step 3. Next, let’s work backwards. How do we solve for F_{x}?
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2}). F_{x} is equivalent to W_{x} in our sketch.
To find the x component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 60^{o}.
β_{1} = β_{2} = 60^{o}
θ_{1} = θ_{2} = 30^{o}If you look at our sketch, you can see that the parallel component of weight (i.e W_{x}) will equal:
W_{x} = (mass)(g)(cosβ_{2})
W_{x} = (30kg)(10m/s^{2})(cos60)
W_{x} = F_{x} = 150 NStep 4. Plug in F_{x} to solve for a_{x}
a_{x} = F_{x} / m
a_{x} = 150 N / 30 kg
a_{x} = 5 m/s^{2}Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Child = 30 kg
 θ = 30^{o}
 g = 10 m/s^{2}
Solving for: child’s acceleration down the slide
Step 2. Next, let’s work backwards. What do we need to solve for the child’s acceleration down the slide?
We need to use the F = m * a equation here.
We know that the net force in the y direction = 0 (i.e the normal force = the y component of the child’s weight). Thus, there there is no acceleration in the y direction. So we are solving for a_{x}, where F_{x} = m * a_{x}
 Rearranging the above: a_{x} = F_{x} / m
 We know m (mass), so we need to determine F_{x}
Step 3. Next, let’s work backwards. How do we solve for F_{x}?
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2}). F_{x} is equivalent to W_{x} in our sketch.
To find the x component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 60^{o}.
β_{1} = β_{2} = 60^{o}
θ_{1} = θ_{2} = 30^{o}If you look at our sketch, you can see that the parallel component of weight (i.e W_{x}) will equal:
W_{x} = (mass)(g)(cosβ_{2})
W_{x} = (30kg)(10m/s^{2})(cos60)
W_{x} = F_{x} = 150 NStep 4. Plug in F_{x} to solve for a_{x}
a_{x} = F_{x} / m
a_{x} = 150 N / 30 kg
a_{x} = 5 m/s^{2}Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 55 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
55. Question
55. Kevin and Betty are holding up a 100 kg piece of furniture with the rope system shown below. What are the tensions in Kevin’s side of the rope (T_{K}), and Betty’s side of the rope (T_{B}), respectively?
Correct / You marked this questionStep 1. We are already provided a sketch. Let’s work backwards. What do we need in order to solve for the tension in Kevin’s rope (T_{K}) and the tension in Betty’s rope (T_{B})?
As the furniture is not moving, we can deduce a few things to start:
1. The tension in the x direction = 0. This implies that T_{Kx} = T_{Bx}
2. The tension in the y direction = 0. This implies that T_{Ky} + T_{By} = mass * gWe now have a system of equations. We can use a little algebra to solve for one of our unknown tensions, and then a little more algebra to use this to solve for the other.
Step 2. Solve for T_{B} in terms of T_{K}
From deduction 1, we can set:
T_{K}cos30 = T_{B} cos 60
T_{K}(^{√3}/_{2}) = ^{TB}/_{2}
T_{B} = √3T_{K}Step 3. Plug our T_{B} value into the equation from deduction 2 to solve for T_{K}:
Step 4. Plug our T_{K} value into the equation from deduction 1 to solve for T_{B}:
T_{B} = √3T_{K}
T_{B} = 500 √3 NIncorrect / You marked this questionStep 1. We are already provided a sketch. Let’s work backwards. What do we need in order to solve for the tension in Kevin’s rope (T_{K}) and the tension in Betty’s rope (T_{B})?
As the furniture is not moving, we can deduce a few things to start:
1. The tension in the x direction = 0. This implies that T_{Kx} = T_{Bx}
2. The tension in the y direction = 0. This implies that T_{Ky} + T_{By} = mass * gWe now have a system of equations. We can use a little algebra to solve for one of our unknown tensions, and then a little more algebra to use this to solve for the other.
Step 2. Solve for T_{B} in terms of T_{K}
From deduction 1, we can set:
T_{K}cos30 = T_{B} cos 60
T_{K}(^{√3}/_{2}) = ^{TB}/_{2}
T_{B} = √3T_{K}Step 3. Plug our T_{B} value into the equation from deduction 2 to solve for T_{K}:
Step 4. Plug our T_{K} value into the equation from deduction 1 to solve for T_{B}:
T_{B} = √3T_{K}
T_{B} = 500 √3 N 
Question 56 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
56. Question
56. Two industrial cranes are holding up a 800 √2 kg box of bricks with the rope system shown below. What are the tensions T_{1} and T_{2} respectively
Correct / You marked this questionStep 1. We are already provided a sketch. Let’s work backwards. What do we need in order to solve for the tension in the ropes for each crane?
As the box of bricks is not moving, we can deduce a few things to start:
1. The tension in the x direction = 0. This implies that T_{1} = T_{2}
2. The tension in the y direction = 0. This implies that T_{1} + T_{2} = mass * gWe now have a system of equations. We can use a little algebra to solve for one of our unknown tensions, and then a little more algebra to use this to solve for the other.
Step 2. Solve for T_{2} in terms of T_{1}
From deduction 1, we can set:
T_{1}cos45 = T_{2}cos 45
T_{1} = T_{2}Step 3. Plug our T_{1} value into the equation from deduction 2 to solve for T_{2}:
T_{1y} + T_{2y} = mass * g
T_{1}sin45 + T_{2}sin 45 = 800 √2 N
2(T_{2}sin 45) = 800 √2 N
T_{2}(^{2}/_{√2}) = 800 √2 N
T_{1} = T_{2} = 800 NIncorrect / You marked this questionStep 1. We are already provided a sketch. Let’s work backwards. What do we need in order to solve for the tension in the ropes for each crane?
As the box of bricks is not moving, we can deduce a few things to start:
1. The tension in the x direction = 0. This implies that T_{1} = T_{2}
2. The tension in the y direction = 0. This implies that T_{1} + T_{2} = mass * gWe now have a system of equations. We can use a little algebra to solve for one of our unknown tensions, and then a little more algebra to use this to solve for the other.
Step 2. Solve for T_{2} in terms of T_{1}
From deduction 1, we can set:
T_{1}cos45 = T_{2}cos 45
T_{1} = T_{2}Step 3. Plug our T_{1} value into the equation from deduction 2 to solve for T_{2}:
T_{1y} + T_{2y} = mass * g
T_{1}sin45 + T_{2}sin 45 = 800 √2 N
2(T_{2}sin 45) = 800 √2 N
T_{2}(^{2}/_{√2}) = 800 √2 N
T_{1} = T_{2} = 800 N 
Question 57 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
57. Question
57. Spiderman shoots two strings of web to suspend a group of bad guys from a roof as shown below. The group of bad guys have a mass of 32 kg. What are the tensions T_{1} and T_{2} respectively?
Correct / You marked this questionStep 1. We are already provided a sketch. Let’s work backwards. What do we need in order to solve for tensions T_{1} and T_{2}?
As the bad guys are suspended (not moving) we can deduce a few things to start:
1. The tension in the x direction = 0. This implies that T_{1x} = T_{2x}
2. The tension in the y direction = 0. This implies that T_{1y} + T_{2y} = mass * gWe now have a system of equations. We can use a little algebra to solve for one of our unknown tensions, and then a little more algebra to use this to solve for the other.
Step 2. Solve for T_{1} in terms of T_{2}
From deduction 1, we can set:
T_{1}cos60 = T_{2}cos30
T_{1}(^{1}/_{2}) = T_{2}(^{√3}/_{2})
T_{1} = √3T_{2}Step 3. Plug our T_{1} value into the equation from deduction 2 to solve for T_{2}:
Step 5. Plug our T_{2} value into the equation from deduction 1 to solve for T_{1}:
T_{1} = √3T_{2}
T_{1} = 160 √3 NIncorrect / You marked this questionStep 1. We are already provided a sketch. Let’s work backwards. What do we need in order to solve for tensions T_{1} and T_{2}?
As the bad guys are suspended (not moving) we can deduce a few things to start:
1. The tension in the x direction = 0. This implies that T_{1x} = T_{2x}
2. The tension in the y direction = 0. This implies that T_{1y} + T_{2y} = mass * gWe now have a system of equations. We can use a little algebra to solve for one of our unknown tensions, and then a little more algebra to use this to solve for the other.
Step 2. Solve for T_{1} in terms of T_{2}
From deduction 1, we can set:
T_{1}cos60 = T_{2}cos30
T_{1}(^{1}/_{2}) = T_{2}(^{√3}/_{2})
T_{1} = √3T_{2}Step 3. Plug our T_{1} value into the equation from deduction 2 to solve for T_{2}:
Step 5. Plug our T_{2} value into the equation from deduction 1 to solve for T_{1}:
T_{1} = √3T_{2}
T_{1} = 160 √3 N 
Question 58 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
58. Question
58. Determine the force attraction due to gravity between a planet of mass 10^{11} kg and comet of mass 10^{3} kg that are √(6.67*10^{3}) meters apart.Correct / You marked this questionStep 1. First, let’s write down our variables and what we are solving for:
 Mass Planet = 10^{11} kg
 Mass Comet = 10^{3} kg
 Distance Between Planet and Comet (d) = √(6.67 x 10^{3}) meters
Solving For: the force attraction (F) due to gravity between the planet and the comet
Step 2. Next, let’s work backwards. What do we need in order to solve for the force attraction due to gravity between the planet and the comet?
You will have to recall the following formula:
Step 3. Plug our given values into the gravitational force formula:
Incorrect / You marked this questionStep 1. First, let’s write down our variables and what we are solving for:
 Mass Planet = 10^{11} kg
 Mass Comet = 10^{3} kg
 Distance Between Planet and Comet (d) = √(6.67 x 10^{3}) meters
Solving For: the force attraction (F) due to gravity between the planet and the comet
Step 2. Next, let’s work backwards. What do we need in order to solve for the force attraction due to gravity between the planet and the comet?
You will have to recall the following formula:
Step 3. Plug our given values into the gravitational force formula:

Question 59 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
59. Question
59. Determine the force attraction due to gravity between a person of mass 100 kg and the Earth of mass 10^{25} kg when the person jumps √6.67 meters off of the ground.Correct / You marked this questionStep 1. First, let’s write down our variables and what we are solving for:
 Mass Earth = 10^{25} kg
 Mass Person = 100 kg
 Distance Between Person and Earth (d) = 6.67 meters
Solving for: the force attraction (F) due to gravity between the Earth and the person
Step 2. Next, let’s work backwards. What do we need in order to solve for the force attraction (F) due to gravity between the Earth and the person?
You will have to recall the following formula:
Step 3. Plug our given values into the gravitational force formula:
Incorrect / You marked this questionStep 1. First, let’s write down our variables and what we are solving for:
 Mass Earth = 10^{25} kg
 Mass Person = 100 kg
 Distance Between Person and Earth (d) = 6.67 meters
Solving for: the force attraction (F) due to gravity between the Earth and the person
Step 2. Next, let’s work backwards. What do we need in order to solve for the force attraction (F) due to gravity between the Earth and the person?
You will have to recall the following formula:
Step 3. Plug our given values into the gravitational force formula:

Question 60 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
60. Question
60. What speed does a rocket of mass 1,000 kg launched into space need to attain to escape Earth’s gravitational attraction? Use mass of Earth is 10^{25} kg and radius of Earth is 6.67 x 10^{6} mCorrect / You marked this questionStep 1. First, let’s write down our variables and what we are solving for:
 Mass Earth = 10^{25} kg
 Mass Rocket = 1000 kg
 Radius of Earth = 6.67 x 10^{6} m
Solving for: escape velocity of the launched rocket
Step 2. Next, let’s work backwards. What do we need in order to solve for the escape velocity of the launched rocket?
You will have to recall the following formula:
Step 3. Plug our given values into the gravitational force formula:
Note 1: if you get a bit stuck on the calculation, look at your answer options. All contain √2 . You can deduce that you will be able to simplify the overall equation to where that something can be square rooted evenly.
Note 2: for escape velocity, make sure you are using the mass of the PLANET and not of the object attempting to escape!
Incorrect / You marked this questionStep 1. First, let’s write down our variables and what we are solving for:
 Mass Earth = 10^{25} kg
 Mass Rocket = 1000 kg
 Radius of Earth = 6.67 x 10^{6} m
Solving for: escape velocity of the launched rocket
Step 2. Next, let’s work backwards. What do we need in order to solve for the escape velocity of the launched rocket?
You will have to recall the following formula:
Step 3. Plug our given values into the gravitational force formula:
Note 1: if you get a bit stuck on the calculation, look at your answer options. All contain √2 . You can deduce that you will be able to simplify the overall equation to where that something can be square rooted evenly.
Note 2: for escape velocity, make sure you are using the mass of the PLANET and not of the object attempting to escape!

Question 61 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
61. Question
61. What speed does a rocket of mass 10 x 10^{11} kg launched into space need to attain to escape the gravitational attraction of Planet X? Use mass of Planet X is 2 x 10^{24} kg and radius of Planet X is 6.67 x 10^{11} mCorrect / You marked this questionStep 1. First, let’s write down our variables and what we are solving for:
 Mass Planet X = 2 x 10^{24} kg
 Mass Rocket = 10 x 10^{11} kg
 Radius of Planet X = 6.67 x 10^{11} m
Solving for: escape velocity of the launched rocket
Step 2. Next, let’s work backwards. What do we need in order to solve for the escape velocity of the launched rocket?
You will have to recall the following formula:
Step 3. Plug our given values into the gravitational force formula:
Note: for escape velocity, make sure you are using the mass of the PLANET and not of the object attempting to escape!
Incorrect / You marked this questionStep 1. First, let’s write down our variables and what we are solving for:
 Mass Planet X = 2 x 10^{24} kg
 Mass Rocket = 10 x 10^{11} kg
 Radius of Planet X = 6.67 x 10^{11} m
Solving for: escape velocity of the launched rocket
Step 2. Next, let’s work backwards. What do we need in order to solve for the escape velocity of the launched rocket?
You will have to recall the following formula:
Step 3. Plug our given values into the gravitational force formula:
Note: for escape velocity, make sure you are using the mass of the PLANET and not of the object attempting to escape!

Question 62 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
62. Question
62. A 10 kg box sits on a flat surface. If the coefficient of static friction is µ_{s} = 0.39, what is the minimum horizontal force that must be applied to the box for it to start moving? Assume gravitational acceleration = 10 m/s^{2}Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box = 10 kg
 g = 10 m/s^{2}
 Coefficient of static friction (µ_{s}) = 0.39
Solving for: the horizontal force required to overcome static friction
Step 2. Next, let’s work backwards. What do we need to solve for the horizontal force required to overcome static friction?
The frictional force acting on the box will equal the Normal Force on the box times the coefficient of static friction.
Thus, to make the box move, the horizontal force applied must exceed the frictional force.
Step 3. Solve for the normal force on the box
Normal Force is the force that a surface exerts on an object that it is in contact with (in this case, the force that the flat surface exerts on the box). This force will be perpendicular to the surface of contact. In this case, as the box is on a flat surface, the normal force will be equal to the box’s weight.
N = m * g = (10 kg) (10 m/s^{2}) = 100 N
Step 4. Solve for frictional force on the box:
Frictional Force (F_{F}) = Normal Force * (µ_{s})
F_{F} = 100 N (0.39) = 39 NThus, in order to make the box move, the horizontal force must be greater than 39 N. Of our answer choices, 40 N is the minimum horizontal force that must be applied to the box for it to start moving
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box = 10 kg
 g = 10 m/s^{2}
 Coefficient of static friction (µ_{s}) = 0.39
Solving for: the horizontal force required to overcome static friction
Step 2. Next, let’s work backwards. What do we need to solve for the horizontal force required to overcome static friction?
The frictional force acting on the box will equal the Normal Force on the box times the coefficient of static friction.
Thus, to make the box move, the horizontal force applied must exceed the frictional force.
Step 3. Solve for the normal force on the box
Normal Force is the force that a surface exerts on an object that it is in contact with (in this case, the force that the flat surface exerts on the box). This force will be perpendicular to the surface of contact. In this case, as the box is on a flat surface, the normal force will be equal to the box’s weight.
N = m * g = (10 kg) (10 m/s^{2}) = 100 N
Step 4. Solve for frictional force on the box:
Frictional Force (F_{F}) = Normal Force * (µ_{s})
F_{F} = 100 N (0.39) = 39 NThus, in order to make the box move, the horizontal force must be greater than 39 N. Of our answer choices, 40 N is the minimum horizontal force that must be applied to the box for it to start moving

Question 63 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
63. Question
63. A box that weighs 10 N box sits on a flat surface. If the coefficient of static friction is µ_{s} = 0.49, what is the minimum horizontal force that must be applied to the box for it to start moving? Assume gravitational acceleration = 10 m/s^{2}Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Weight Box = 10 N
 g = 10 m/s^{2}
 coefficient of static friction (µ_{s}) = 0.49
Solving for: the horizontal force required to overcome static friction
Step 2. Next, let’s work backwards. What do we need to solve for the horizontal force required to overcome static friction?
The frictional force acting on the box will equal the Normal Force on the box times the coefficient of static friction.
Thus, to make the box move, the horizontal force applied must exceed the frictional force.
Step 3. Solve for the normal force on the box.
Normal Force is the force that a surface exerts on an object that it is in contact with (in this case, the force that the flat surface exerts on the box). This force will be perpendicular to the surface of contact. In this case, as the box is on a flat surface, the normal force will be equal to the box’s weight.
N = W_{box} = 10 N
Note: you are already given the box’s weight, so you don’t need to multiply this by g.
Step 4. Solve for frictional force on the box:
Frictional Force (F_{F}) = Normal Force (µ_{s})
F_{F} = 10 N (0.49) = 4.9 NThus, in order to make the box move, the horizontal force must be greater than 4.9 N. Of our answer choices, 5 N is the minimum horizontal force that must be applied to the box for it to start moving
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Weight Box = 10 N
 g = 10 m/s^{2}
 coefficient of static friction (µ_{s}) = 0.49
Solving for: the horizontal force required to overcome static friction
Step 2. Next, let’s work backwards. What do we need to solve for the horizontal force required to overcome static friction?
The frictional force acting on the box will equal the Normal Force on the box times the coefficient of static friction.
Thus, to make the box move, the horizontal force applied must exceed the frictional force.
Step 3. Solve for the normal force on the box.
Normal Force is the force that a surface exerts on an object that it is in contact with (in this case, the force that the flat surface exerts on the box). This force will be perpendicular to the surface of contact. In this case, as the box is on a flat surface, the normal force will be equal to the box’s weight.
N = W_{box} = 10 N
Note: you are already given the box’s weight, so you don’t need to multiply this by g.
Step 4. Solve for frictional force on the box:
Frictional Force (F_{F}) = Normal Force (µ_{s})
F_{F} = 10 N (0.49) = 4.9 NThus, in order to make the box move, the horizontal force must be greater than 4.9 N. Of our answer choices, 5 N is the minimum horizontal force that must be applied to the box for it to start moving

Question 64 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
64. Question
64. A 50 kg box is placed on a declined hill. The hill forms a 30^{o} angle with the horizontal. If the coefficient of static friction is µ_{s} = ^{√3}/_{5}, what applied force must be exceeded to cause the box to move downhill? Assume gravitational acceleration = 10 m/s^{2}Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box = 50 kg
 θ = 30^{o}
 g = 10 m/s^{2}
Solving for: the applied force that must be exceeded to cause the box to move downhill
Step 2. Next, let’s work backwards. What do we need to solve for the horizontal force required to overcome static friction?
The frictional force acting on the box will equal the Normal Force on the box times the coefficient of static friction.
Thus, to make the box move, the parallel force applied must exceed the frictional force.
Step 3. Next, let’s work backwards. What do we need to solve for the normal force (N) on the box?
Normal Force is the force that a surface exerts on an object that it is in contact with (in this case, the force that the hill exerts on the box). This force will be perpendicular to the surface of contact (in this case, the slant of the hill).
Thus, to solve for the Normal Force, we must solve for the perpendicular component of the object’s weight. In this case, we have termed this “W_{y}”
Step 4. Next, let’s work backwards. How do we solve for W_{y}?
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2})
To find the y component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 60^{o}.
β_{1} = β_{2} = 60^{o}.
θ_{1} = θ_{2} = 30^{o}If you look at our sketch, you can see that the perpendicular component of weight (i.e W_{y}) will equal:
W_{y} = (mass)(g)(sinβ_{2})
W_{y} = (50kg)(10m/s^{2})(sin60)
W_{y} = N = 250√3 NStep 5. Solve for frictional force on the box:
Frictional Force (F_{F}) = Normal Force (µ_{s})
F_{F} = (250√3 N)(^{√3}/_{5}) = 150 NThus, in order to make the box move, the horizontal force must be greater than 39 N. Of our answer choices, the applied force must exceed 150 N to cause the box to start moving downhill.
Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box = 50 kg
 θ = 30^{o}
 g = 10 m/s^{2}
Solving for: the applied force that must be exceeded to cause the box to move downhill
Step 2. Next, let’s work backwards. What do we need to solve for the horizontal force required to overcome static friction?
The frictional force acting on the box will equal the Normal Force on the box times the coefficient of static friction.
Thus, to make the box move, the parallel force applied must exceed the frictional force.
Step 3. Next, let’s work backwards. What do we need to solve for the normal force (N) on the box?
Normal Force is the force that a surface exerts on an object that it is in contact with (in this case, the force that the hill exerts on the box). This force will be perpendicular to the surface of contact (in this case, the slant of the hill).
Thus, to solve for the Normal Force, we must solve for the perpendicular component of the object’s weight. In this case, we have termed this “W_{y}”
Step 4. Next, let’s work backwards. How do we solve for W_{y}?
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2})
To find the y component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 60^{o}.
β_{1} = β_{2} = 60^{o}.
θ_{1} = θ_{2} = 30^{o}If you look at our sketch, you can see that the perpendicular component of weight (i.e W_{y}) will equal:
W_{y} = (mass)(g)(sinβ_{2})
W_{y} = (50kg)(10m/s^{2})(sin60)
W_{y} = N = 250√3 NStep 5. Solve for frictional force on the box:
Frictional Force (F_{F}) = Normal Force (µ_{s})
F_{F} = (250√3 N)(^{√3}/_{5}) = 150 NThus, in order to make the box move, the horizontal force must be greater than 39 N. Of our answer choices, the applied force must exceed 150 N to cause the box to start moving downhill.
Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 65 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
65. Question
65. A 10 kg child slides down a slide. The slide forms a 60^{o} angle with the horizontal. The child started from rest, and accelerates at a rate of 2.5 m/s^{2}. Determine the coefficient of kinetic friction for the slide. Assume gravitational acceleration = 10 m/s^{2}Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Child = 10 kg
 θ = 60^{o}
 g = 10 m/s^{2}
 a = 2.5 m/s^{2}
Solving for: the coefficient of kinetic friction for the slide
Step 2. Next, let’s work backwards. What do we need to solve for the coefficient of kinetic friction for the slide?
We know that the equation regarding friction is as follows:
We are given mass and acceleration. Thus, we need to solve for normal force (F_{N}) and friction force (F_{F}) to be able to solve for µ_{k}
Step 3. Next, let’s work backwards. What do we need to solve for the normal force (F_{N}) on the box?
Normal Force is the force that a surface exerts on an object that it is in contact with (in this case, the force that the slide exerts on the child). This force will be perpendicular to the surface of contact (in this case, the slant of the slide).
Thus, to solve for the Normal Force, we must solve for the perpendicular component of the child’s weight. In this case, we have termed this “W_{y}”
Step 4. Next, let’s work backwards. How do we solve for W_{y}?
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2})
To find the y component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 30^{o}.
β_{1} = β_{2} = 30^{o}.
θ_{1} = θ_{2} = 60^{o}If you look at our sketch, you can see that the perpendicular component of weight (i.e W_{y}) will equal:
W_{y} = (mass)(g)(sinβ_{2})
W_{y} = (10 kg)(10 m/s^{2})(sin30)
W_{y} = F_{N} = 50 NStep 5. Next, solve for the friction force (F_{F}) on the box.
We need to use the F = m * a equation here.
F_{F} = m * a
F_{F} = (10 kg)(2.5 m/s^{2})
F_{F} = 25 NStep 6. Last, solve for the coefficient of kinetic friction for the slide.
F_{F} = µ_{k}F_{N}
25 N = µ_{k}(50 N)
µ_{k} = 0.5Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Child = 10 kg
 θ = 60^{o}
 g = 10 m/s^{2}
 a = 2.5 m/s^{2}
Solving for: the coefficient of kinetic friction for the slide
Step 2. Next, let’s work backwards. What do we need to solve for the coefficient of kinetic friction for the slide?
We know that the equation regarding friction is as follows:
We are given mass and acceleration. Thus, we need to solve for normal force (F_{N}) and friction force (F_{F}) to be able to solve for µ_{k}
Step 3. Next, let’s work backwards. What do we need to solve for the normal force (F_{N}) on the box?
Normal Force is the force that a surface exerts on an object that it is in contact with (in this case, the force that the slide exerts on the child). This force will be perpendicular to the surface of contact (in this case, the slant of the slide).
Thus, to solve for the Normal Force, we must solve for the perpendicular component of the child’s weight. In this case, we have termed this “W_{y}”
Step 4. Next, let’s work backwards. How do we solve for W_{y}?
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2})
To find the y component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 30^{o}.
β_{1} = β_{2} = 30^{o}.
θ_{1} = θ_{2} = 60^{o}If you look at our sketch, you can see that the perpendicular component of weight (i.e W_{y}) will equal:
W_{y} = (mass)(g)(sinβ_{2})
W_{y} = (10 kg)(10 m/s^{2})(sin30)
W_{y} = F_{N} = 50 NStep 5. Next, solve for the friction force (F_{F}) on the box.
We need to use the F = m * a equation here.
F_{F} = m * a
F_{F} = (10 kg)(2.5 m/s^{2})
F_{F} = 25 NStep 6. Last, solve for the coefficient of kinetic friction for the slide.
F_{F} = µ_{k}F_{N}
25 N = µ_{k}(50 N)
µ_{k} = 0.5Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 66 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
66. Question
66. In a packaging factory, a 15 kg box slides down a slide. The slide forms a 45^{o} angle with the horizontal. The box started from rest, and accelerates at a rate of √2 m/s^{2}. Determine the coefficient of kinetic friction for the slide. Assume gravitational acceleration = 10 m/s^{2}Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box = 15 kg
 θ = 45^{o}
 g = 10 m/s^{2}
 a = √2 m/s^{2}
Solving for: the coefficient of kinetic friction for the slide
Step 2. Next, let’s work backwards. What do we need to solve for the coefficient of kinetic friction for the slide?
We know that the equation regarding friction is as follows:
We are given mass and acceleration. Thus, we need to solve for normal force (F_{N}) and friction force (F_{F}) to be able to solve for µ_{k}
Step 3. Next, let’s work backwards. What do we need to solve for the normal force (F_{N}) on the box?
Normal Force is the force that a surface exerts on an object that it is in contact with (in this case, the force that the slide exerts on the box). This force will be perpendicular to the surface of contact (in this case, the slant of the slide).
Thus, to solve for the Normal Force, we must solve for the perpendicular component of the object’s weight. In this case, we have termed this “W_{y}”
Step 4. Next, let’s work backwards. How do we solve for W_{y}?
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2})
To find the y component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 45^{o}.
β_{1} = β_{2} = 45^{o}.
θ_{1} = θ_{2} = 45^{o}If you look at our sketch, you can see that the perpendicular component of weight (i.e W_{y}) will equal:
W_{y} = (mass)(g)(sinβ_{2})
W_{y} = (15 kg)(10 m/s^{2})(sin45)
W_{y} = F_{N} = ^{150}/_{√2} NStep 5. Next, solve for the friction force (F_{F}) on the box.
We need to use the F = m * a equation here.
F_{F} = m * a
F_{F} = (15 kg)(√2 m/s^{2})
F_{F} = 15√2 NStep 6. Last, solve for the coefficient of kinetic friction for the slide.
F_{F} = µ_{k}F_{N}
15√2 N = µ_{k}(^{150}/_{√2} N)
µ_{k} = 0.2Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box = 15 kg
 θ = 45^{o}
 g = 10 m/s^{2}
 a = √2 m/s^{2}
Solving for: the coefficient of kinetic friction for the slide
Step 2. Next, let’s work backwards. What do we need to solve for the coefficient of kinetic friction for the slide?
We know that the equation regarding friction is as follows:
We are given mass and acceleration. Thus, we need to solve for normal force (F_{N}) and friction force (F_{F}) to be able to solve for µ_{k}
Step 3. Next, let’s work backwards. What do we need to solve for the normal force (F_{N}) on the box?
Normal Force is the force that a surface exerts on an object that it is in contact with (in this case, the force that the slide exerts on the box). This force will be perpendicular to the surface of contact (in this case, the slant of the slide).
Thus, to solve for the Normal Force, we must solve for the perpendicular component of the object’s weight. In this case, we have termed this “W_{y}”
Step 4. Next, let’s work backwards. How do we solve for W_{y}?
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2})
To find the y component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 45^{o}.
β_{1} = β_{2} = 45^{o}.
θ_{1} = θ_{2} = 45^{o}If you look at our sketch, you can see that the perpendicular component of weight (i.e W_{y}) will equal:
W_{y} = (mass)(g)(sinβ_{2})
W_{y} = (15 kg)(10 m/s^{2})(sin45)
W_{y} = F_{N} = ^{150}/_{√2} NStep 5. Next, solve for the friction force (F_{F}) on the box.
We need to use the F = m * a equation here.
F_{F} = m * a
F_{F} = (15 kg)(√2 m/s^{2})
F_{F} = 15√2 NStep 6. Last, solve for the coefficient of kinetic friction for the slide.
F_{F} = µ_{k}F_{N}
15√2 N = µ_{k}(^{150}/_{√2} N)
µ_{k} = 0.2Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 67 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
67. Question
67. In an airport, a 30 kg suitcase slides down a conveyor belt. The conveyor belt forms a 30^{o} angle with the horizontal. The suitcase started from rest, and accelerates at a rate of 3√3 m/s^{2}. Determine the coefficient of kinetic friction for the slide. Assume gravitational acceleration = 10 m/s^{2}Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box = 15 kg
 θ = 30^{o}
 g = 10 m/s^{2}
 a = 3√3 m/s^{2}
Solving for: the coefficient of kinetic friction for the conveyor belt
Step 2. Next, let’s work backwards. What do we need to solve for the coefficient of kinetic friction for the slide?
We know that the equation regarding friction is as follows:
We are given mass and acceleration. Thus, we need to solve for normal force (F_{N}) and friction force (F_{F}) to be able to solve for µ_{k}.
Step 3. Next, let’s work backwards. What do we need to solve for the normal force (F_{N}) on the box?
Normal Force is the force that a surface exerts on an object that it is in contact with (in this case, the force that the conveyor belt exerts on the suitcase). This force will be perpendicular to the surface of contact (in this case, the slant of the conveyor belt).
Thus, to solve for the Normal Force, we must solve for the perpendicular component of the suitcase’s weight. In this case, we have termed this “W_{y}”
Step 4. Next, let’s work backwards. How do we solve for W_{y}?
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2})
To find the y component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 60^{o}.
β_{1} = β_{2} = 60^{o}.
θ_{1} = θ_{2} = 30^{o}If you look at our sketch, you can see that the perpendicular component of weight (i.e W_{y}) will equal:
W_{y} = (mass)(g)(sinβ_{2})
W_{y} = (30 kg)(10 m/s^{2})(sin60)
W_{y} = F_{N} = 150√3 NStep 5. Next, solve for the friction force (F_{F}) on the box.
We need to use the F = m * a equation here.
F_{F} = m * a
F_{F} = (30 kg)(3√3 m/s^{2})
F_{F} = 90√3 NStep 6. Last, solve for the coefficient of kinetic friction for the slide.
F_{F} = µ_{k}F_{N}
90√3 N = µ_{k}(150√3 N)
µ_{k} = 0.6Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box = 15 kg
 θ = 30^{o}
 g = 10 m/s^{2}
 a = 3√3 m/s^{2}
Solving for: the coefficient of kinetic friction for the conveyor belt
Step 2. Next, let’s work backwards. What do we need to solve for the coefficient of kinetic friction for the slide?
We know that the equation regarding friction is as follows:
We are given mass and acceleration. Thus, we need to solve for normal force (F_{N}) and friction force (F_{F}) to be able to solve for µ_{k}.
Step 3. Next, let’s work backwards. What do we need to solve for the normal force (F_{N}) on the box?
Normal Force is the force that a surface exerts on an object that it is in contact with (in this case, the force that the conveyor belt exerts on the suitcase). This force will be perpendicular to the surface of contact (in this case, the slant of the conveyor belt).
Thus, to solve for the Normal Force, we must solve for the perpendicular component of the suitcase’s weight. In this case, we have termed this “W_{y}”
Step 4. Next, let’s work backwards. How do we solve for W_{y}?
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2})
To find the y component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 60^{o}.
β_{1} = β_{2} = 60^{o}.
θ_{1} = θ_{2} = 30^{o}If you look at our sketch, you can see that the perpendicular component of weight (i.e W_{y}) will equal:
W_{y} = (mass)(g)(sinβ_{2})
W_{y} = (30 kg)(10 m/s^{2})(sin60)
W_{y} = F_{N} = 150√3 NStep 5. Next, solve for the friction force (F_{F}) on the box.
We need to use the F = m * a equation here.
F_{F} = m * a
F_{F} = (30 kg)(3√3 m/s^{2})
F_{F} = 90√3 NStep 6. Last, solve for the coefficient of kinetic friction for the slide.
F_{F} = µ_{k}F_{N}
90√3 N = µ_{k}(150√3 N)
µ_{k} = 0.6Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 68 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
68. Question
68. Two people stand in an elevator. Their combined mass is 150 kg. They press the “23” button, and the elevator begins to accelerate upwards at 5 m/s^{2}. What is their combined weight in the accelerating elevator?Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch and what we are solving for:
 Combined Mass = 150 kg
 Upward Acceleration (a) = 5 m/s^{2}
 Gravitational Acceleration (g) = 10 m/s^{2}
Solving for: the duo’s combined weight in ascending elevator
Step 2. Let’s work backwards. What do we need to solve for their combined weight in ascending elevator?
Remember that weight is a force, and for an object at rest, F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2})
Thus, the duo’s weight in the elevator at rest = 150 kg * (10 m/s^{2}) = 1500 N
When you are riding in an elevator accelerating upwards, the normal force and force of acceleration are in the same direction. Thus, you weight will be greater.
Thus, for an elevator accelerating upwards:
F_{weight} = mass * (g + acceleration of elevator)
In this case:
F_{weight} = 150kg * (10 m/s^{2} + 5 m/s^{2})
F_{weight} = 2250 NIncorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch and what we are solving for:
 Combined Mass = 150 kg
 Upward Acceleration (a) = 5 m/s^{2}
 Gravitational Acceleration (g) = 10 m/s^{2}
Solving for: the duo’s combined weight in ascending elevator
Step 2. Let’s work backwards. What do we need to solve for their combined weight in ascending elevator?
Remember that weight is a force, and for an object at rest, F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2})
Thus, the duo’s weight in the elevator at rest = 150 kg * (10 m/s^{2}) = 1500 N
When you are riding in an elevator accelerating upwards, the normal force and force of acceleration are in the same direction. Thus, you weight will be greater.
Thus, for an elevator accelerating upwards:
F_{weight} = mass * (g + acceleration of elevator)
In this case:
F_{weight} = 150kg * (10 m/s^{2} + 5 m/s^{2})
F_{weight} = 2250 N 
Question 69 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
69. Question
69. A person stands in an elevator. In the elevator at rest, they weigh 500 N. They press the “45” button, and the elevator begins to accelerate upwards at 2 m/s^{2}. How much does the person weight in the ascending elevator? Assume gravitational acceleration = 10 m/s^{2}Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Weight = 500 N
 Upward Acceleration (a) = 5 m/s^{2}
 Gravitational Acceleration (g) = 10 m/s^{2}
Solving for: the person’s weight in ascending elevator
Step 2. Let’s work backwards. What do we need to solve for their combined weight in ascending elevator?
Remember that weight is a force, and for an object at rest, F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2})
We are given that the person’s weight in the elevator at rest is rest = 500 N. Thus, to solve for the person’s mass:
F_{weight} = mass * g
500 N = mass * (10 m/s_{2})
mass = 50 kgWhen you are riding in an elevator accelerating upwards, the normal force and force of acceleration are in the same direction. Thus, your weight will be greater.
Thus, for an elevator accelerating upwards:
F_{weight} = mass * (g + acceleration of elevator)
In this case:
F_{weight} = 50 kg * (10 m/s^{2} + 2 m/s^{2})
F_{weight} = 600 NIncorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Weight = 500 N
 Upward Acceleration (a) = 5 m/s^{2}
 Gravitational Acceleration (g) = 10 m/s^{2}
Solving for: the person’s weight in ascending elevator
Step 2. Let’s work backwards. What do we need to solve for their combined weight in ascending elevator?
Remember that weight is a force, and for an object at rest, F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2})
We are given that the person’s weight in the elevator at rest is rest = 500 N. Thus, to solve for the person’s mass:
F_{weight} = mass * g
500 N = mass * (10 m/s_{2})
mass = 50 kgWhen you are riding in an elevator accelerating upwards, the normal force and force of acceleration are in the same direction. Thus, your weight will be greater.
Thus, for an elevator accelerating upwards:
F_{weight} = mass * (g + acceleration of elevator)
In this case:
F_{weight} = 50 kg * (10 m/s^{2} + 2 m/s^{2})
F_{weight} = 600 N 
Question 70 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
70. Question
70. You stand in an elevator on the 1st floor of a building. In the elevator at rest, you weigh 300 N. You press the “10” button, and the elevator begins to rise upwards at a constant velocity. What is your weight in the ascending elevator? Assume gravitational acceleration = 10 m/s^{2}Correct / You marked this questionLet’s draw a quick sketch:
 If an elevator is ascending at a constant velocity, that means the acceleration of the elevator = 0
 Your weight (m*g) will be equal in magnitude to the normal force of the elevator pushing up against you (+m*g) (though in opposite directions).
 Thus, when the elevator is not accelerating (either at rest, or moving at a constant velocity EITHER UPWARDS OR DOWNWARDS) the weight of the object in the elevator will not change.
 In this case, your weight will remain 300 N. Note that weight is not measured in kg, so you can immediately discount any options with this unit.
Incorrect / You marked this questionLet’s draw a quick sketch:
 If an elevator is ascending at a constant velocity, that means the acceleration of the elevator = 0
 Your weight (m*g) will be equal in magnitude to the normal force of the elevator pushing up against you (+m*g) (though in opposite directions).
 Thus, when the elevator is not accelerating (either at rest, or moving at a constant velocity EITHER UPWARDS OR DOWNWARDS) the weight of the object in the elevator will not change.
 In this case, your weight will remain 300 N. Note that weight is not measured in kg, so you can immediately discount any options with this unit.

Question 71 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
71. Question
71. You stand in an elevator on the 1^{st} floor of a building. In the elevator at rest, you weigh 200 N. You press the “10” button, and the elevator begins to rise upwards. As it nears the 10^{th} floor, it begins to decelerate at a rate of 2 m/s^{2}. What is your weight at this moment, as the elevator nears the 10^{th} floor? Assume gravitational acceleration = 10 m/s^{2}Correct / You marked this questionLet’s draw a quick sketch:
 At this moment, the acceleration of the elevator is downwards, and the direction of movement of the elevator is upwards.
 Thus, the force of acceleration and the normal force are in OPPOSITE directions. This implies that your weight will be less than your original weight.
 In this case, your weight will be less than 200 N. Note that weight is not measured in kg, so you can immediately discount any options with this unit.
Incorrect / You marked this questionLet’s draw a quick sketch:
 At this moment, the acceleration of the elevator is downwards, and the direction of movement of the elevator is upwards.
 Thus, the force of acceleration and the normal force are in OPPOSITE directions. This implies that your weight will be less than your original weight.
 In this case, your weight will be less than 200 N. Note that weight is not measured in kg, so you can immediately discount any options with this unit.

Question 72 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
72. Question
72. A 200 kg man pushes a 100 kg couch with a force of 50 N. The coefficient of kinetic friction (µ_{k}) = 0.77. What is the force the couch exerts on the man?Correct / You marked this questionThis is a conceptual question. Newton’s Third Law says that for every action, there is an equal and opposite reaction. This implies that if the man pushes on the couch with a force of 50 N, the couch pushes back on the man with an equal force of 50 N.
Incorrect / You marked this questionThis is a conceptual question. Newton’s Third Law says that for every action, there is an equal and opposite reaction. This implies that if the man pushes on the couch with a force of 50 N, the couch pushes back on the man with an equal force of 50 N.

Question 73 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
73. Question
73. A phone company launches a satellite (“Satellite 1”) into orbit around the Earth. The 350 kg satellite orbits the Earth at a velocity of 500 m/s. If the phone company launches a 700 kg satellite (“Satellite 2”) to orbit at the same distance from Earth, what will the ratio of the orbital velocity of Satellite 1 to Satellite 2? For mass of Earth, use: 10^{25} kg For radius of Earth, use: 6.67 x 10^{6} mCorrect / You marked this questionStep 1. First, let’s write down our variables and what we are solving for:
 Mass Earth = 10^{25} kg
 Mass Satellite 1 = 350 kg
 Mass Satellite 2 = 700 kg
 Radius of Earth = 6.67 x 10^{6} m
Solving for: the ratio of the orbital velocity of satellite 1 to satellite 2
Step 2. Next, let’s work backwards. What do we need to solve for the ratio of the velocity of Satellite 1 versus Satellite 2?
You will have to recall the following formula:
We know that we only need to solve for the ratio of the velocity of Satellite 1 versus Satellite 2. In most cases, this means we won’t need to fully solve for each velocity, but can merely compare the velocities based on changing one variable and holding the rest constant.
Step 3. Compare the velocities based on changing one variable and holding the rest constant.
We can clearly see that the velocities are equal, i.e. in a ratio of 1:1.
It would be easy to get this problem wrong if we had used the mass of the satellites for “m.” It is important to become familiar with the explanations of each variable in the far right column of the equations sheet!
Incorrect / You marked this questionStep 1. First, let’s write down our variables and what we are solving for:
 Mass Earth = 10^{25} kg
 Mass Satellite 1 = 350 kg
 Mass Satellite 2 = 700 kg
 Radius of Earth = 6.67 x 10^{6} m
Solving for: the ratio of the orbital velocity of satellite 1 to satellite 2
Step 2. Next, let’s work backwards. What do we need to solve for the ratio of the velocity of Satellite 1 versus Satellite 2?
You will have to recall the following formula:
We know that we only need to solve for the ratio of the velocity of Satellite 1 versus Satellite 2. In most cases, this means we won’t need to fully solve for each velocity, but can merely compare the velocities based on changing one variable and holding the rest constant.
Step 3. Compare the velocities based on changing one variable and holding the rest constant.
We can clearly see that the velocities are equal, i.e. in a ratio of 1:1.
It would be easy to get this problem wrong if we had used the mass of the satellites for “m.” It is important to become familiar with the explanations of each variable in the far right column of the equations sheet!

Question 74 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
74. Question
74. NASA launches a satellite (“Satellite 1”) into orbit around the Earth. The 1000 kg satellite orbits the Earth at a velocity of 500 m/s. If NASA launches a 1000 kg satellite (“Satellite 2”) into orbit twice as far from Earth’s center, what will the ratio of the orbital velocity of Satellite 1 to Satellite 2? For mass of Earth, use: 10^{25} kg For radius of Earth, use: 6.67 x 10^{6} mCorrect / You marked this questionStep 1. First, let’s write down our variables and what we are solving for:
 Mass Earth = 10^{25} kg
 Mass Satellite 1 = 1000 kg
 Mass Satellite 2 = 1000 kg
 Radius of Earth = 6.67 x 10^{6} m
Solving for: the ratio of the orbital velocity of Satellite 1 to Satellite 2
Step 2. Next, let’s work backwards. What do we need to solve for the ratio of the velocity of Satellite 1 versus Satellite 2?
You will have to recall the following formula:
We know that we only need to solve for the ratio of the velocity of Satellite 1 versus Satellite 2. In most cases, this means we won’t need to fully solve for each velocity, but can merely compare the velocities based on changing one variable and holding the rest constant.
Step 3. Compare the velocities based on changing one variable and holding the rest constant.
Thus, the velocities are in a ratio of 1:(^{1}/_{√2}). If this is difficult for you to see, remember that:
Incorrect / You marked this questionStep 1. First, let’s write down our variables and what we are solving for:
 Mass Earth = 10^{25} kg
 Mass Satellite 1 = 1000 kg
 Mass Satellite 2 = 1000 kg
 Radius of Earth = 6.67 x 10^{6} m
Solving for: the ratio of the orbital velocity of Satellite 1 to Satellite 2
Step 2. Next, let’s work backwards. What do we need to solve for the ratio of the velocity of Satellite 1 versus Satellite 2?
You will have to recall the following formula:
We know that we only need to solve for the ratio of the velocity of Satellite 1 versus Satellite 2. In most cases, this means we won’t need to fully solve for each velocity, but can merely compare the velocities based on changing one variable and holding the rest constant.
Step 3. Compare the velocities based on changing one variable and holding the rest constant.
Thus, the velocities are in a ratio of 1:(^{1}/_{√2}). If this is difficult for you to see, remember that:

Question 75 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
75. Question
75. Which of the answer choices are units of acceleration?Correct / You marked this questionAcceleration is the rate of change of velocity. We know that velocity is typically in units of m/s. Thus, and object’s acceleration will typically be given in units of m/s per second, otherwise denoted as m/s^{2}
Incorrect / You marked this questionAcceleration is the rate of change of velocity. We know that velocity is typically in units of m/s. Thus, and object’s acceleration will typically be given in units of m/s per second, otherwise denoted as m/s^{2}

Question 76 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
76. Question
76. Starting from rest, a race car driver that presses as hard as he can on the gas from the starting gate will have:Correct / You marked this questionAcceleration is the rate of change of velocity. When the race car driver comes out from the start, the car’s acceleration will be high, as the car’s velocity is changing very quickly. His speed will be slow, however, since he is starting initially from rest.
Incorrect / You marked this questionAcceleration is the rate of change of velocity. When the race car driver comes out from the start, the car’s acceleration will be high, as the car’s velocity is changing very quickly. His speed will be slow, however, since he is starting initially from rest.

Question 77 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
77. Question
77. A race car driver gets a flat tire. He chugs along at constant but not impressive pace, as other cars zoom past him. His car has:Correct / You marked this questionAcceleration is the rate of change of velocity. As the car is moving at a constant pace, its acceleration will be zero. As the car is moving slowly with a flat tire, its speed will be slow.
Incorrect / You marked this questionAcceleration is the rate of change of velocity. As the car is moving at a constant pace, its acceleration will be zero. As the car is moving slowly with a flat tire, its speed will be slow.

Question 78 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
78. Question
78. A race car driver zooming along makes a final push to get past the driver in first place. They are neck and neck, and then he stomps on the gas to push ahead. As he is making the pass, his car has:Correct / You marked this questionAcceleration is the rate of change of velocity. As the car moves faster in order to pass the first place car, his car is accelerating. You can tell by the fact that he is in first place that his speed must be fast!
Incorrect / You marked this questionAcceleration is the rate of change of velocity. As the car moves faster in order to pass the first place car, his car is accelerating. You can tell by the fact that he is in first place that his speed must be fast!

Question 79 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
79. Question
79. What is the ratio of the force of gravitational between the planet and the comet in scenario A versus scenario B:
A) A planet of mass 1 x 10^{11} kg and comet of mass 10^{3} kg that are 5 x 10^{3} meters apart.
B) A planet of mass 2 x 10^{11} kg and comet of mass 10^{3} kg that are 10 x 10^{3} meters apart.Correct / You marked this questionStep 1. What do we need in order to solve for the force attraction due to gravity between the planet and the comet in each scenario?
You will have to recall the following formula:
We know that we only need to solve for the ratio of the force of attraction in Scenario A versus Scenario B. In most cases, this means we won’t need to fully solve for each attraction force, but can merely compare the forces based on changing one or two variables and holding the rest constant.
Step 2. Compare the forces based on changing mass of planet and distance apart and holding the rest constant.
Thus, the forces are in a ratio of 1:0.5 otherwise denoted as 2:1.
Incorrect / You marked this questionStep 1. What do we need in order to solve for the force attraction due to gravity between the planet and the comet in each scenario?
You will have to recall the following formula:
We know that we only need to solve for the ratio of the force of attraction in Scenario A versus Scenario B. In most cases, this means we won’t need to fully solve for each attraction force, but can merely compare the forces based on changing one or two variables and holding the rest constant.
Step 2. Compare the forces based on changing mass of planet and distance apart and holding the rest constant.
Thus, the forces are in a ratio of 1:0.5 otherwise denoted as 2:1.

Question 80 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
80. Question
80. A 30 kg (Box A) and a 50 kg (Box B) box are attached by a rope on a frictionless surface. A man pulls on Box A with 80 N of force. What is the force of tension in the rope?Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box A = 30 kg
 Mass Box B = 50 kg
 Force Man Pulling = 80 N
Solving for: the force of tension in the rope
Step 2. Next, let’s work backwards. What do we need to solve for the force of tension in the rope?
Looking at the entire system, we know that the two objects will move with the same velocity and acceleration, as they are attached.
Thus, we can use the F_{net} = ma equation to solve for the acceleration of both boxes.
From there, we can look more specifically at the rope and Box B system, to determine the force of tension in the rope.
Step 3. Solve for the acceleration of both boxes.
F_{net} = ma
80 N = (30 kg + 50 kg)(a)
a = 1 m/s^{2}Step 4. Solve for the force of tension in the rope and Box B system
F_{net} = Force of Tension = ma
F_{net} = Force of Tension = 50 kg * 1 m/s^{2}
Force of Tension = 50 NIncorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box A = 30 kg
 Mass Box B = 50 kg
 Force Man Pulling = 80 N
Solving for: the force of tension in the rope
Step 2. Next, let’s work backwards. What do we need to solve for the force of tension in the rope?
Looking at the entire system, we know that the two objects will move with the same velocity and acceleration, as they are attached.
Thus, we can use the F_{net} = ma equation to solve for the acceleration of both boxes.
From there, we can look more specifically at the rope and Box B system, to determine the force of tension in the rope.
Step 3. Solve for the acceleration of both boxes.
F_{net} = ma
80 N = (30 kg + 50 kg)(a)
a = 1 m/s^{2}Step 4. Solve for the force of tension in the rope and Box B system
F_{net} = Force of Tension = ma
F_{net} = Force of Tension = 50 kg * 1 m/s^{2}
Force of Tension = 50 N 
Question 81 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
81. Question
81. A 15 kg (Box A) and a 45 kg (Box B) box are attached by a rope on a frictionless surface. A man pulls on Box A with 180 N of force. What is the force of tension in the rope?Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box A = 15 kg
 Mass Box B = 45 kg
 Force Man Pulling = 180 N
Solving for: the force of tension in the rope
Step 2. Next, let’s work backwards. What do we need to solve for the force of tension in the rope?
Looking at the entire system, we know that the two objects will move with the same velocity and acceleration, as they are attached.
Thus, we can use the F_{net} = ma equation to solve for the acceleration of both boxes.
From there, we can look more specifically at the rope and Box B system, to determine the force of tension in the rope.
Step 3. Solve for the acceleration of both boxes.
F_{net} = ma
180 N = (15 kg + 45 kg)(a)
a = 3 m/s^{2}Step 4. Solve for the force of tension in the rope and Box B system
F_{net} = Force of Tension = ma
F_{net} = Force of Tension = 45 kg * 3 m/s^{2}
Force of Tension = 135 NIncorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box A = 15 kg
 Mass Box B = 45 kg
 Force Man Pulling = 180 N
Solving for: the force of tension in the rope
Step 2. Next, let’s work backwards. What do we need to solve for the force of tension in the rope?
Looking at the entire system, we know that the two objects will move with the same velocity and acceleration, as they are attached.
Thus, we can use the F_{net} = ma equation to solve for the acceleration of both boxes.
From there, we can look more specifically at the rope and Box B system, to determine the force of tension in the rope.
Step 3. Solve for the acceleration of both boxes.
F_{net} = ma
180 N = (15 kg + 45 kg)(a)
a = 3 m/s^{2}Step 4. Solve for the force of tension in the rope and Box B system
F_{net} = Force of Tension = ma
F_{net} = Force of Tension = 45 kg * 3 m/s^{2}
Force of Tension = 135 N 
Question 82 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
82. Question
82. You have three boxes that you plan to return to the store, all lined up in a row (Box A = 3 kg, then Box B = 4 kg, then Box C = 5 kg). You push them across your frictionless floor (pushing from Box A) towards the door with force of 24 N. What is the magnitude of the contact force between Box B and Box C?Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box A = 3 kg
 Mass Box B = 4 kg
 Mass Box C = 5 kg
 Pushing Force = 24 N
Solving for: the magnitude of the contact force between Box B and Box C
Step 2. Next, let’s work backwards. What do we need to solve for the force of tension in the rope?
Looking at the entire system, we know that the three will move with the same velocity and acceleration, as they are all in contact.
Thus, we can use the F_{net} = ma equation to solve for the acceleration of all three boxes.
From there, we can look more specifically at the contact force between Box B and Box C.
Step 3. Solve for the acceleration of all three boxes.
F_{net} = ma
24 N = (3kg + 4kg + 5kg) (a)
a = 2 m/s^{2}Step 4. Solve for the contact force between box B and C
F_{contact} = ma
F_{contact} = 5 kg (2 m/s^{2})
F_{contact} = 10 NImportant: think intuitively about why we use the mass of Box C to determine the contact force in this problem. What do we really mean by contact force here?
You are pushing these three boxes along. The contact force between Box B and Box C is the force with which Box C is pushing back on on Box B (and ultimately, making it harder for you to push your boxes!)
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box A = 3 kg
 Mass Box B = 4 kg
 Mass Box C = 5 kg
 Pushing Force = 24 N
Solving for: the magnitude of the contact force between Box B and Box C
Step 2. Next, let’s work backwards. What do we need to solve for the force of tension in the rope?
Looking at the entire system, we know that the three will move with the same velocity and acceleration, as they are all in contact.
Thus, we can use the F_{net} = ma equation to solve for the acceleration of all three boxes.
From there, we can look more specifically at the contact force between Box B and Box C.
Step 3. Solve for the acceleration of all three boxes.
F_{net} = ma
24 N = (3kg + 4kg + 5kg) (a)
a = 2 m/s^{2}Step 4. Solve for the contact force between box B and C
F_{contact} = ma
F_{contact} = 5 kg (2 m/s^{2})
F_{contact} = 10 NImportant: think intuitively about why we use the mass of Box C to determine the contact force in this problem. What do we really mean by contact force here?
You are pushing these three boxes along. The contact force between Box B and Box C is the force with which Box C is pushing back on on Box B (and ultimately, making it harder for you to push your boxes!)

Question 83 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
83. Question
83. You have three boxes that you plan to return to the store, all lined up in a row (Box A = 3 kg, then Box B = 4 kg, then Box C = 5 kg). You push them across your frictionless floor (pushing from Box A) towards the door with force of 24 N. What is the magnitude of the contact force, the resisting force, between Box A and Box B?Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box A = 3 kg
 Mass Box B = 4 kg
 Mass Box C = 5 kg
 Pushing Force = 24 N
Solving for: the magnitude of the contact force between Box B and Box C
Step 2. Next, let’s work backwards. What do we need to solve for the force of tension in the rope?
Looking at the entire system, we know that the three will move with the same velocity and acceleration, as they are all in contact. This is key concept to keep in mind with a moving system.
Thus, we can use the F_{net} = ma equation to solve for the acceleration of both all three boxes.
From there, we can look more specifically at the contact force between Box A and Box B.
Step 3. Solve for the acceleration of all three boxes.
F_{net} = ma
24 N = (3kg + 4kg + 5kg) (a)
a = 2 m/s^{2}Step 4. Solve for the contact force between box A and B
F_{contact} = ma
F_{contact} = (4 kg +5 kg)(2 m/s^{2})
F_{contact} = 18 NImportant: think intuitively about why we use the mass combined mass of Box B and C to determine the contact force in this problem. What do we really mean by contact force here?
You are pushing these three boxes along. The contact force between Box A and Box B is the force with which Box B and C is pushing back on Box A (and ultimately, making it harder for you to push your boxes!)
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box A = 3 kg
 Mass Box B = 4 kg
 Mass Box C = 5 kg
 Pushing Force = 24 N
Solving for: the magnitude of the contact force between Box B and Box C
Step 2. Next, let’s work backwards. What do we need to solve for the force of tension in the rope?
Looking at the entire system, we know that the three will move with the same velocity and acceleration, as they are all in contact. This is key concept to keep in mind with a moving system.
Thus, we can use the F_{net} = ma equation to solve for the acceleration of both all three boxes.
From there, we can look more specifically at the contact force between Box A and Box B.
Step 3. Solve for the acceleration of all three boxes.
F_{net} = ma
24 N = (3kg + 4kg + 5kg) (a)
a = 2 m/s^{2}Step 4. Solve for the contact force between box A and B
F_{contact} = ma
F_{contact} = (4 kg +5 kg)(2 m/s^{2})
F_{contact} = 18 NImportant: think intuitively about why we use the mass combined mass of Box B and C to determine the contact force in this problem. What do we really mean by contact force here?
You are pushing these three boxes along. The contact force between Box A and Box B is the force with which Box B and C is pushing back on Box A (and ultimately, making it harder for you to push your boxes!)

Question 84 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
84. Question
84. Two boxes are attached by a rope as shown in the diagram below. Box A slides on a frictionless table. Box B hangs over the edge of the table via a frictionless pulley. Determine the tension in the rope. Use gravitational acceleration (g) = 10 m/s^{2}
Correct / You marked this questionStep 1. We know a few things from the get go.
 As the two boxes are attached by the rope, they will have the same acceleration (a).
 The tension in the rope is uniform throughout the rope.
Let’s look at the net force on each box to solve for acceleration (a) in each case. You’ll see momentarily how we can use the fact that acceleration is equal for both boxes in our favor.
Step 2. Compare net force Box A versus Box B
For Box A:
F_{net} = m_{A}a
T = 5 kg (a)
a = ^{T}/_{5 kg}Note, as T (tension in the rope) is the only force acting on Box A, we were able to substitute T for F_{net}
For Box B:
F_{net} = force due to gravity MINUS tension in the rope
m_{B}g – T = m_{B} (a)
a = g – (^{T}/_{m}_{B})
a = g – (^{T}/_{10 kg})This is probably the most difficult step in the problem. Here are the two things you must realize to get this right:
1. We set F_{net} as force of gravity minus the tension in the rope (versus the other way around) because:
 The tension in the rope is pulling Box A towards the right, and Box B upwards (essentially in opposite directions)
 The force of gravity is pulling Box B downwards, which is the same direction the tension in the rope is pulling Box A
The arrows on the rope in the picture should help you see this.
2. We set the force of gravity as m_{B}g (versus just g) because F = m * a.
Step 3. Set acceleration from our first equation equal to acceleration from our second equation to solve for T. We can do this as they are moving as one system.
Incorrect / You marked this questionStep 1. We know a few things from the get go.
 As the two boxes are attached by the rope, they will have the same acceleration (a).
 The tension in the rope is uniform throughout the rope.
Let’s look at the net force on each box to solve for acceleration (a) in each case. You’ll see momentarily how we can use the fact that acceleration is equal for both boxes in our favor.
Step 2. Compare net force Box A versus Box B
For Box A:
F_{net} = m_{A}a
T = 5 kg (a)
a = ^{T}/_{5 kg}Note, as T (tension in the rope) is the only force acting on Box A, we were able to substitute T for F_{net}
For Box B:
F_{net} = force due to gravity MINUS tension in the rope
m_{B}g – T = m_{B} (a)
a = g – (^{T}/_{m}_{B})
a = g – (^{T}/_{10 kg})This is probably the most difficult step in the problem. Here are the two things you must realize to get this right:
1. We set F_{net} as force of gravity minus the tension in the rope (versus the other way around) because:
 The tension in the rope is pulling Box A towards the right, and Box B upwards (essentially in opposite directions)
 The force of gravity is pulling Box B downwards, which is the same direction the tension in the rope is pulling Box A
The arrows on the rope in the picture should help you see this.
2. We set the force of gravity as m_{B}g (versus just g) because F = m * a.
Step 3. Set acceleration from our first equation equal to acceleration from our second equation to solve for T. We can do this as they are moving as one system.

Question 85 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
85. Question
85. Two boxes are attached by a rope as shown in the diagram below. Box A slides on a frictionless table. Box B hangs over the edge of the table via a frictionless pulley. Determine the tension in the rope. Use gravitational acceleration (g) = 10 m/s^{2}
Correct / You marked this questionStep 1. We know a few things from the get go.
 As the two boxes are attached by the rope, they will have the same acceleration (a).
 The tension in the rope is uniform throughout the rope.
Let’s look at the net force on each box to solve for acceleration (a) in each case. You’ll see momentarily how we can use the fact that acceleration is equal for both boxes in our favor.
Step 2. Compare net force Box A versus Box B
For Box A:
F_{net} = m_{A}a
T = 30 kg (a)
a = ^{T}/_{30 kg}Note, as T (tension in the rope) is the only force acting on Box A, we were able to substitute T for F_{net}
For Box B:
F_{net} = force due to gravity MINUS tension in the rope
m_{B}g – T = m_{B} (a)
a = g – (^{T}/_{m}_{B})
a = g – (^{T}/_{6 kg})This is probably the most difficult step in the problem. Here are the two things you must realize to get this right:
1. We set F_{net} as force of gravity minus the tension in the rope (versus the other way around) because:
 The tension in the rope is pulling Box A towards the right, and Box B upwards (essentially in opposite directions)
 The force of gravity is pulling Box B downwards, which is the same direction the tension in the rope is pulling Box A
The arrows on the rope in the picture should help you see this.
2. We set the force of gravity as m_{B}g (versus just g) because F = m * a.
Step 3. Set acceleration from our first equation equal to acceleration from our second equation to solve for T
Incorrect / You marked this questionStep 1. We know a few things from the get go.
 As the two boxes are attached by the rope, they will have the same acceleration (a).
 The tension in the rope is uniform throughout the rope.
Let’s look at the net force on each box to solve for acceleration (a) in each case. You’ll see momentarily how we can use the fact that acceleration is equal for both boxes in our favor.
Step 2. Compare net force Box A versus Box B
For Box A:
F_{net} = m_{A}a
T = 30 kg (a)
a = ^{T}/_{30 kg}Note, as T (tension in the rope) is the only force acting on Box A, we were able to substitute T for F_{net}
For Box B:
F_{net} = force due to gravity MINUS tension in the rope
m_{B}g – T = m_{B} (a)
a = g – (^{T}/_{m}_{B})
a = g – (^{T}/_{6 kg})This is probably the most difficult step in the problem. Here are the two things you must realize to get this right:
1. We set F_{net} as force of gravity minus the tension in the rope (versus the other way around) because:
 The tension in the rope is pulling Box A towards the right, and Box B upwards (essentially in opposite directions)
 The force of gravity is pulling Box B downwards, which is the same direction the tension in the rope is pulling Box A
The arrows on the rope in the picture should help you see this.
2. We set the force of gravity as m_{B}g (versus just g) because F = m * a.
Step 3. Set acceleration from our first equation equal to acceleration from our second equation to solve for T

Question 86 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
86. Question
86. Box A and Box B are attached by a rope over a frictionless pulley as shown in the diagram below. Determine the tension in the rope. Use gravitational acceleration (g) = 10 m/s^{2}
Correct / You marked this questionStep 1. We know a few things from the get go.
 As the two boxes are attached by the rope, they will have the same acceleration (a), but in opposite directions. The downward acceleration of Box A will have the same magnitude as the upward acceleration of Box B.
 The tension in the rope is uniform throughout the rope.
Let’s look at the net force on each box to solve for acceleration (a) in each case. You’ll see momentarily how we can use the fact that acceleration is equal for both boxes in our favor.
Step 2. Compare net force Box A versus Box B
For Box A:
F_{net} = force due to gravity MINUS tension in the rope
m_{A}g – T = m_{A} (a)
a = g – (^{T}/_{m}_{A})
a = g – (^{T}/_{20 kg})For Box B:
F_{net} = tension in the rope MINUS force due to gravity
T – m_{B}g = m_{B}(a)
a = (^{T}/_{m}_{B}) – g
a = (^{T}/_{5 kg}) – gThis is probably the most difficult step in the problem. Here are a few things you must realize to get this right:
1. For Box A, we set F_{net} as force of gravity minus the tension in the rope because the Box is moving in the direction that gravity is pulling it.
2. For Box B, we set F_{net} as the tension in the rope minus the acceleration of gravity, because Box B is moving upwards (the direction that Box A is pulling it) against gravity. The arrows on the rope in the picture should help you see this.
3. We set the force of gravity as m_{A}g or m_{B}g (versus just g) because F = m * a.
Step 3. Set acceleration from our first equation equal to acceleration from our second equation to solve for T
Incorrect / You marked this questionStep 1. We know a few things from the get go.
 As the two boxes are attached by the rope, they will have the same acceleration (a), but in opposite directions. The downward acceleration of Box A will have the same magnitude as the upward acceleration of Box B.
 The tension in the rope is uniform throughout the rope.
Let’s look at the net force on each box to solve for acceleration (a) in each case. You’ll see momentarily how we can use the fact that acceleration is equal for both boxes in our favor.
Step 2. Compare net force Box A versus Box B
For Box A:
F_{net} = force due to gravity MINUS tension in the rope
m_{A}g – T = m_{A} (a)
a = g – (^{T}/_{m}_{A})
a = g – (^{T}/_{20 kg})For Box B:
F_{net} = tension in the rope MINUS force due to gravity
T – m_{B}g = m_{B}(a)
a = (^{T}/_{m}_{B}) – g
a = (^{T}/_{5 kg}) – gThis is probably the most difficult step in the problem. Here are a few things you must realize to get this right:
1. For Box A, we set F_{net} as force of gravity minus the tension in the rope because the Box is moving in the direction that gravity is pulling it.
2. For Box B, we set F_{net} as the tension in the rope minus the acceleration of gravity, because Box B is moving upwards (the direction that Box A is pulling it) against gravity. The arrows on the rope in the picture should help you see this.
3. We set the force of gravity as m_{A}g or m_{B}g (versus just g) because F = m * a.
Step 3. Set acceleration from our first equation equal to acceleration from our second equation to solve for T

Question 87 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
87. Question
87. Box A and Box B are attached by a rope over a frictionless pulley as shown in the diagram below. Determine the tension in the rope. Use gravitational acceleration (g) = 10 m/s^{2}
Correct / You marked this questionStep 1. We know a few things from the get go.
 As Box A has the same mass as Box B, the acceleration of both Box A and Box B will be equal to zero.
 The tension in the rope is uniform throughout the rope.
The tension in the rope will be one component of the net force on either box. Let’s look at the net force on Box A to solve for tension in the rope.
Step 2. Write out net force Box A equation, then solve for tension in the rope
For Box A:
Another way to solve this problem would be to realize that the net force on Box A F_{net} = 0 (because the box is not moving). Thus, the magnitude of the force due to gravity = tension in the rope.
m_{A}g = T
(7 kg)(10 m/s^{2}) = T
T = 70 NIncorrect / You marked this questionStep 1. We know a few things from the get go.
 As Box A has the same mass as Box B, the acceleration of both Box A and Box B will be equal to zero.
 The tension in the rope is uniform throughout the rope.
The tension in the rope will be one component of the net force on either box. Let’s look at the net force on Box A to solve for tension in the rope.
Step 2. Write out net force Box A equation, then solve for tension in the rope
For Box A:
Another way to solve this problem would be to realize that the net force on Box A F_{net} = 0 (because the box is not moving). Thus, the magnitude of the force due to gravity = tension in the rope.
m_{A}g = T
(7 kg)(10 m/s^{2}) = T
T = 70 N 
Question 88 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
88. Question
88. You step on a scale on Earth, and it reads 40 kg. How much would you weigh on Planet X, which is 4x the mass of Earth, and 4x the radius of Earth? Use gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionStep 1. First, let’s write down our variables and what we are solving for:
 Weight Earth = 40 kg
 Mass Planet X = 4x Earth
 Radius Planet X = 4x Earth
Solving for: your weight on Planet X
Step 2. Next, let’s work backwards. What do we need to solve for your weight on Planet X?
This calls for our gravitational force between two bodies equation.
If we solve for the gravitational force on Earth, then the gravitational force on Planet X, we can compare the ratio of the two.
Step 3. Plug in our known values to the gravitational force equation to determine the gravitational force for each planet. Once we determine the ratio of gravitational force on Planet X versus Earth, we can determine the specific weight difference.
Note: your mass on Earth will be exactly the same on Earth and Planet X. Mass does not change based on location!
Step 4. Compare the ratio of the two gravitational forces.
 F(Planet X) will be (4x value of Earth due to mass) / (16x value of Earth – due to radius)
 ^{4}/_{16} = ^{1}/_{4}. This implies the ratio of your weight on Planet X to Earth will be 1:4.
 Your weight on Earth = 40 kg * 10 m/s^{2} = 400 N.
 Thus, your weight on Planet X = 100 N
Incorrect / You marked this questionStep 1. First, let’s write down our variables and what we are solving for:
 Weight Earth = 40 kg
 Mass Planet X = 4x Earth
 Radius Planet X = 4x Earth
Solving for: your weight on Planet X
Step 2. Next, let’s work backwards. What do we need to solve for your weight on Planet X?
This calls for our gravitational force between two bodies equation.
If we solve for the gravitational force on Earth, then the gravitational force on Planet X, we can compare the ratio of the two.
Step 3. Plug in our known values to the gravitational force equation to determine the gravitational force for each planet. Once we determine the ratio of gravitational force on Planet X versus Earth, we can determine the specific weight difference.
Note: your mass on Earth will be exactly the same on Earth and Planet X. Mass does not change based on location!
Step 4. Compare the ratio of the two gravitational forces.
 F(Planet X) will be (4x value of Earth due to mass) / (16x value of Earth – due to radius)
 ^{4}/_{16} = ^{1}/_{4}. This implies the ratio of your weight on Planet X to Earth will be 1:4.
 Your weight on Earth = 40 kg * 10 m/s^{2} = 400 N.
 Thus, your weight on Planet X = 100 N

Question 89 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
89. Question
89. Which of the following units are equivalent to Newtons? Use gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionWe know that Newtons are the unit of Force, and F_{net} = m*a.
 Mass is typically in units of kg.
 Acceleration is typically in units of m/s^{2}
 Thus, Newtons are equivalent to kg * m/s^{2}
Incorrect / You marked this questionWe know that Newtons are the unit of Force, and F_{net} = m*a.
 Mass is typically in units of kg.
 Acceleration is typically in units of m/s^{2}
 Thus, Newtons are equivalent to kg * m/s^{2}

Question 90 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
90. Question
90. The tortoise and the hare are in a race. The tortoise gets 50 m ahead of the hare before the hare decides to start. The tortoise moves at a steady pace of 5 m/s. The hare starts going and accelerates at 2 m/s^{2}. The race is 100 m total. By how many meters does the winner win the race?Correct / You marked this questionStep 1. Let’s draw a quick sketch:
Step 2. What will we need to solve for the distance by which the winner wins by?
Let’s use our displacement equation to determine the distance traveled by each animal as a function of time.
We know that in general: d = v_{i}t + ^{1}/_{2}at^{2}
For the tortoise:
d = 50 m + v_{i}t + ^{1}/_{2}at^{2}
d = 50 m + v_{i}t
d = 50 m + (5 m/s) tFor the hare:
d = v_{i}t + ^{1}/_{2}at^{2}
d = ^{1}/_{2}at^{2}Step 3. Let’s find out how many seconds it will take each animal to finish the race.
For the tortoise:
d = 50 m + (5 m/s) t
100 m = 50m + (5m/s) t
t = 10 sFor the hare:
d = ^{1}/_{2}at^{2}
100 m = ^{1}/_{2}(2 m/s^{2})t^{2}
100 m = 1 m/s^{2} * t^{2}
t = 10 sThus, both animals will finish the race at the same time.
Incorrect / You marked this questionStep 1. Let’s draw a quick sketch:
Step 2. What will we need to solve for the distance by which the winner wins by?
Let’s use our displacement equation to determine the distance traveled by each animal as a function of time.
We know that in general: d = v_{i}t + ^{1}/_{2}at^{2}
For the tortoise:
d = 50 m + v_{i}t + ^{1}/_{2}at^{2}
d = 50 m + v_{i}t
d = 50 m + (5 m/s) tFor the hare:
d = v_{i}t + ^{1}/_{2}at^{2}
d = ^{1}/_{2}at^{2}Step 3. Let’s find out how many seconds it will take each animal to finish the race.
For the tortoise:
d = 50 m + (5 m/s) t
100 m = 50m + (5m/s) t
t = 10 sFor the hare:
d = ^{1}/_{2}at^{2}
100 m = ^{1}/_{2}(2 m/s^{2})t^{2}
100 m = 1 m/s^{2} * t^{2}
t = 10 sThus, both animals will finish the race at the same time.

Question 91 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
91. Question
91. A 10 kg child prepares to slide down a slide. The slide forms a 60^{o} angle with the horizontal. Assuming the child starts from rest, what is the minimum coefficient of static friction that will prevent the child from moving down the slide? Assume gravitational acceleration = 10 m/s^{2}Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Child = 10 kg
 θ = 60^{o}
 g = 10 m/s^{2}
Solving for: the minimum coefficient of static friction that will prevent the child from sliding down the slide
Step 2. Next, let’s work backwards. What do we need to solve for the minimum coefficient of static friction that will prevent the child from sliding down the slide?
To prevent the child from sliding down the slide, the force due to friction must be greater than or equal to the gravitational force pulling the child down the slide.
We know that the equation regarding friction is as follows:
Mathematically, we express this as: F_{F} = µ_{S}F_{N} ≥ W_{x}
Thus, we need to solve for normal force (F_{N}) and friction force (W_{x}) to be able to solve for µ_{S}
Step 3. Next, let’s work backwards. What do we need to solve for the normal force (F_{N}) on the box?
Normal Force is the force that a surface exerts on an object that it is in contact with (in this case, the force that the slide exerts on the child). This force will be perpendicular to the surface of contact (in this case, the slant of the slide).
Thus, to solve for the Normal Force, we must solve for the perpendicular component of the child’s weight. In this case, we have termed this “W_{y}”Step 4. Next, let’s work backwards. How do we solve for W_{y}?
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2})
To find the y component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 30^{o}.
β_{1} = β_{2} = 30^{o}
θ_{1} = θ_{2} = 60^{o}If you look at our sketch, you can see that the perpendicular component of weight (i.e W_{y}) will equal:
W_{y} = (mass)(g)(sinβ_{2})
W_{y} = (10 kg)(10 m/s^{2})(sin30)
W_{y} = F_{N} = 50 NStep 5. Next, solve for the force W_{x} on the box.
We need to use the F = m * a equation here.
W_{x} = (mass)(g)(cosβ_{2})
W_{x} = (10 kg)(10 m/s^{2})(cos30)
W_{x} = 50 √3 NStep 6. Last, solve for the minimum coefficient of static for the slide.
µ_{S}F_{N} ≥ W_{x}
µ_{S} * 50 N ≥ 50 √3 N
µ_{S} ≥ √3Note: the coefficient of friction (either kinetic or static) CAN be greater than 1. The coefficient of friction demonstrates relationship between the force of friction and the normal force. A frictional coefficient greater than 1 implies that the force of friction is greater than the normal force.
Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Child = 10 kg
 θ = 60^{o}
 g = 10 m/s^{2}
Solving for: the minimum coefficient of static friction that will prevent the child from sliding down the slide
Step 2. Next, let’s work backwards. What do we need to solve for the minimum coefficient of static friction that will prevent the child from sliding down the slide?
To prevent the child from sliding down the slide, the force due to friction must be greater than or equal to the gravitational force pulling the child down the slide.
We know that the equation regarding friction is as follows:
Mathematically, we express this as: F_{F} = µ_{S}F_{N} ≥ W_{x}
Thus, we need to solve for normal force (F_{N}) and friction force (W_{x}) to be able to solve for µ_{S}
Step 3. Next, let’s work backwards. What do we need to solve for the normal force (F_{N}) on the box?
Normal Force is the force that a surface exerts on an object that it is in contact with (in this case, the force that the slide exerts on the child). This force will be perpendicular to the surface of contact (in this case, the slant of the slide).
Thus, to solve for the Normal Force, we must solve for the perpendicular component of the child’s weight. In this case, we have termed this “W_{y}”Step 4. Next, let’s work backwards. How do we solve for W_{y}?
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2})
To find the y component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 30^{o}.
β_{1} = β_{2} = 30^{o}
θ_{1} = θ_{2} = 60^{o}If you look at our sketch, you can see that the perpendicular component of weight (i.e W_{y}) will equal:
W_{y} = (mass)(g)(sinβ_{2})
W_{y} = (10 kg)(10 m/s^{2})(sin30)
W_{y} = F_{N} = 50 NStep 5. Next, solve for the force W_{x} on the box.
We need to use the F = m * a equation here.
W_{x} = (mass)(g)(cosβ_{2})
W_{x} = (10 kg)(10 m/s^{2})(cos30)
W_{x} = 50 √3 NStep 6. Last, solve for the minimum coefficient of static for the slide.
µ_{S}F_{N} ≥ W_{x}
µ_{S} * 50 N ≥ 50 √3 N
µ_{S} ≥ √3Note: the coefficient of friction (either kinetic or static) CAN be greater than 1. The coefficient of friction demonstrates relationship between the force of friction and the normal force. A frictional coefficient greater than 1 implies that the force of friction is greater than the normal force.
Knowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 92 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
92. Question
92. A box is in contact with a flat surface. No net force acts upon the box. Which of the following must be true?Correct / You marked this questionLet’s look at each answer option in turn:
[A] The box has zero acceleration.
This is the correct answer. We know: F_{net} = ma. Thus, if F_{net} = 0, then a must equal zero. This is Newton’s 2^{nd} Law of Motion.
[B] The box is not moving
There could be a scenario where no net force is acting up on the box, and the box is moving with constant velocity.
[C] The force of kinetic friction is equal to the force of static friction
By definition, you either have kinetic friction (when an object is in motion) or static friction (when an object is stationary). To compare the two thus does not make sense.
[D] The box has nonzero kinetic energy
If the box is stationary, the box would have zero kinetic energy.
[E] The box’s displacement is zero
In part B, we established that there could be a scenario where no net force is acting up on the box, and the box is moving with constant velocity. Thus, the box’s displacement from its starting point would be nonzero.
Incorrect / You marked this questionLet’s look at each answer option in turn:
[A] The box has zero acceleration.
This is the correct answer. We know: F_{net} = ma. Thus, if F_{net} = 0, then a must equal zero. This is Newton’s 2^{nd} Law of Motion.
[B] The box is not moving
There could be a scenario where no net force is acting up on the box, and the box is moving with constant velocity.
[C] The force of kinetic friction is equal to the force of static friction
By definition, you either have kinetic friction (when an object is in motion) or static friction (when an object is stationary). To compare the two thus does not make sense.
[D] The box has nonzero kinetic energy
If the box is stationary, the box would have zero kinetic energy.
[E] The box’s displacement is zero
In part B, we established that there could be a scenario where no net force is acting up on the box, and the box is moving with constant velocity. Thus, the box’s displacement from its starting point would be nonzero.

Question 93 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
93. Question
93. A bullet is shot horizontally out of a window (Bullet A) at the same time an identical bullet is dropped from the same window (Bullet B). Which one will hit the ground first?Correct / You marked this questionThis is a fundamental principle that you are expected to understand for the OAT regarding projectile motion. If two objects fall from the same height, REGARDLESS of the horizontal component of their motion, (and also, regardless of their masses) it will take the exact same amount of time for each to reach the ground.
This is of course, neglecting air resistance, which you can safely assume is the case if not told otherwise.
Incorrect / You marked this questionThis is a fundamental principle that you are expected to understand for the OAT regarding projectile motion. If two objects fall from the same height, REGARDLESS of the horizontal component of their motion, (and also, regardless of their masses) it will take the exact same amount of time for each to reach the ground.
This is of course, neglecting air resistance, which you can safely assume is the case if not told otherwise.

Question 94 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
94. Question
94. A 20 kg bullet is shot upwards out of window at 30^{o} relative to the horizontal (Bullet A) at the same time an 80 kg bullet is shot out of the same window at the same angle (Bullet B). Both are shot with the same initial velocity. Which one will hit the ground first?Correct / You marked this questionThis is a fundamental principle that you are expected to understand for the OAT regarding projectile motion. If two objects start from the same height, and have the same vertical component of motion (in this scenario, being shot at the same angle with the same velocity), REGARDLESS of their masses (and also, the horizontal component of their motion) it will take the exact same amount of time for each to reach the ground.
This is of course, neglecting air resistance, which you can safely assume is the case if not told otherwise.
Incorrect / You marked this questionThis is a fundamental principle that you are expected to understand for the OAT regarding projectile motion. If two objects start from the same height, and have the same vertical component of motion (in this scenario, being shot at the same angle with the same velocity), REGARDLESS of their masses (and also, the horizontal component of their motion) it will take the exact same amount of time for each to reach the ground.
This is of course, neglecting air resistance, which you can safely assume is the case if not told otherwise.

Question 95 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
95. Question
95. You and your brother are racing boxes across your frictionless floor. Starting from rest, you push your 10 kg box for 10 s with a force of 10 N. What is the velocity of your box at time t = 10 seconds?Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box = 10 kg
 Pushing Force = 10 N
 Time = 10 seconds
Solving for: the velocity of your box at time t = 10 seconds (v_{f})
Step 2. Next, let’s work backwards. What do we need to solve for the velocity of your box at time t = 10 seconds?
We know that: F_{net} = ma
10 N = 10 kg * a
a = 1 m/s^{2}From here, we can use another equation from our sheet to solve for the velocity at time t = 10 seconds (v_{f})
Step 3. Solve for v_{f}
v_{f} = v_{i} + at
v_{f} = 0 m/s + (1 m/s^{2})(10 s)
v_{f} = 10 m/sIncorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Box = 10 kg
 Pushing Force = 10 N
 Time = 10 seconds
Solving for: the velocity of your box at time t = 10 seconds (v_{f})
Step 2. Next, let’s work backwards. What do we need to solve for the velocity of your box at time t = 10 seconds?
We know that: F_{net} = ma
10 N = 10 kg * a
a = 1 m/s^{2}From here, we can use another equation from our sheet to solve for the velocity at time t = 10 seconds (v_{f})
Step 3. Solve for v_{f}
v_{f} = v_{i} + at
v_{f} = 0 m/s + (1 m/s^{2})(10 s)
v_{f} = 10 m/s 
Question 96 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
96. Question
96. A 1000 kg rocket launches from the ground. After 15 m, it has attained a velocity of 30 m/s. What is the magnitude of the upward force was required to propel the rocket as described? Assume gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Rocket = 1000 kg
 Distance Traveled (d) = 15 m
 Initial Velocity (v_{i}) = 0 m/s
 Final Velocity (v_{f}) = 30 m/s
Solving for: the magnitude of the upward force was required to propel the rocket
Step 2. Next, let’s work backwards. What do we need to solve for the magnitude of the upward force was required to propel the rocket?
We know that: F_{net} = ma
 There are two forces acting on the rocket. The downward force of gravity, and the upward force of the engines.
 In order to propel the rocket as described, the upward force of the engines must overcome the force of gravity.
 So, the magnitude of upward force required will be equal to the magnitude of the force of gravity on the rocket + the remaining force required to cause the velocity change.
Step 3. Solve for the force of gravity on the rocket:
F_{GRAVITY} = ma = mg
F_{GRAVITY} = 1000 kg * 10 m/s^{2}
F_{GRAVITY} = 10,000 NIf you think about it, any upward force ≤ 10,000 N would not cause the rocket to move upwards at all. You can think about this as the “threshold” force for upward movement.
Step 4. Solve for the remaining force required to cause the velocity change:
Once we apply 10,000 N of force, we have “accounted for” the force of gravity. At this point, we solve for the remaining upward force required to cause the velocity change in the same way we would if the rocket were moving horizontally with no friction.
i.e. by using the following equation:
v_{f}^{2} = v_{i}^{2} + 2ad
(30 m/s)^{2} = 0 + 2(a)(15 m)
900 m^{2}/s^{2} = (30 m)(a)
a = 30 m/s^{2}We know that: F = ma
F = 1000 kg * 30 m/s^{2}
F = 30,000 NThus, the magnitude of the TOTAL upward force required to propel the rocket will be equal to:
(F_{GRAVITY} = 10,000 N) + (F = 30,000 N)
F_{TOTAL} = 40,000 NIncorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Rocket = 1000 kg
 Distance Traveled (d) = 15 m
 Initial Velocity (v_{i}) = 0 m/s
 Final Velocity (v_{f}) = 30 m/s
Solving for: the magnitude of the upward force was required to propel the rocket
Step 2. Next, let’s work backwards. What do we need to solve for the magnitude of the upward force was required to propel the rocket?
We know that: F_{net} = ma
 There are two forces acting on the rocket. The downward force of gravity, and the upward force of the engines.
 In order to propel the rocket as described, the upward force of the engines must overcome the force of gravity.
 So, the magnitude of upward force required will be equal to the magnitude of the force of gravity on the rocket + the remaining force required to cause the velocity change.
Step 3. Solve for the force of gravity on the rocket:
F_{GRAVITY} = ma = mg
F_{GRAVITY} = 1000 kg * 10 m/s^{2}
F_{GRAVITY} = 10,000 NIf you think about it, any upward force ≤ 10,000 N would not cause the rocket to move upwards at all. You can think about this as the “threshold” force for upward movement.
Step 4. Solve for the remaining force required to cause the velocity change:
Once we apply 10,000 N of force, we have “accounted for” the force of gravity. At this point, we solve for the remaining upward force required to cause the velocity change in the same way we would if the rocket were moving horizontally with no friction.
i.e. by using the following equation:
v_{f}^{2} = v_{i}^{2} + 2ad
(30 m/s)^{2} = 0 + 2(a)(15 m)
900 m^{2}/s^{2} = (30 m)(a)
a = 30 m/s^{2}We know that: F = ma
F = 1000 kg * 30 m/s^{2}
F = 30,000 NThus, the magnitude of the TOTAL upward force required to propel the rocket will be equal to:
(F_{GRAVITY} = 10,000 N) + (F = 30,000 N)
F_{TOTAL} = 40,000 N 
Question 97 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
97. Question
97. Box A and Box B are attached by a rope over a frictionless pulley as shown in the diagram below. They are released at the same time, at the same height. Determine how long it will take for Box A to fall 12 m to hit the ground. Use gravitational acceleration (g) = 10 m/s^{2}
Correct / You marked this questionStep 1. We know a few things from the get go.
 As the two boxes are attached by the rope, they will have the same acceleration (a), but in opposite directions. The downward acceleration of Box A will have the same magnitude as the upward acceleration of Box B.
 The tension in the rope is uniform throughout the rope.
Let’s look at the net force on each box to solve for acceleration (a) in each case. You’ll see momentarily how we can use the fact that acceleration is equal for both boxes in our favor.
Step 2. Compare net force Box A versus Box B
For Box A:
F_{net} = force due to gravity MINUS tension in the rope
m_{A}g – T = m_{A} (a)
a = g – (^{T}/_{m}_{A})
a = g – (^{T}/_{20 kg})For Box B:
F_{net} = tension in the rope MINUS force due to gravity
T – m_{B}g = m_{B}(a)
a = (^{T}/_{m}_{B}) – g
a = (^{T}/_{5 kg}) – gThis is probably the most difficult step in the problem. Here are a few things you must realize to get this right:
1. For Box A, we set F_{net} as force of gravity minus the tension in the rope because the Box is moving in the direction that gravity is pulling it.
2. For Box B, we set F_{net} as the tension in the rope minus the acceleration of gravity, because Box B is moving upwards (the direction that Box A is pulling it) against gravity. The arrows on the rope in the picture should help you see this.
3. We set the force of gravity as m_{A}g or m_{B}g (versus just g) because F = m * a.
Step 3. Set acceleration from our first equation equal to acceleration from our second equation to solve for T
Step 4. Use Box A acceleration to solve for acceleration (a)
a = g – (^{T}/^{20 kg})
a = 10 m/s^{2} – (^{80 N}/_{20 kg})
a = 6 m/s^{2}Note that this acceleration is in the downward direction, which in this scenario, we have chosen to define as positive for simplicity.
Step 5. From here, we can use the displacement equation for the time it takes for the box to fall:
d = v_{i}t + ^{1}/_{2}at^{2}
12 m = 0 + ^{1}/_{2}(6 m/s^{2})t^{2}
12 m = 3t^{2}
t = 2 secondsIncorrect / You marked this questionStep 1. We know a few things from the get go.
 As the two boxes are attached by the rope, they will have the same acceleration (a), but in opposite directions. The downward acceleration of Box A will have the same magnitude as the upward acceleration of Box B.
 The tension in the rope is uniform throughout the rope.
Let’s look at the net force on each box to solve for acceleration (a) in each case. You’ll see momentarily how we can use the fact that acceleration is equal for both boxes in our favor.
Step 2. Compare net force Box A versus Box B
For Box A:
F_{net} = force due to gravity MINUS tension in the rope
m_{A}g – T = m_{A} (a)
a = g – (^{T}/_{m}_{A})
a = g – (^{T}/_{20 kg})For Box B:
F_{net} = tension in the rope MINUS force due to gravity
T – m_{B}g = m_{B}(a)
a = (^{T}/_{m}_{B}) – g
a = (^{T}/_{5 kg}) – gThis is probably the most difficult step in the problem. Here are a few things you must realize to get this right:
1. For Box A, we set F_{net} as force of gravity minus the tension in the rope because the Box is moving in the direction that gravity is pulling it.
2. For Box B, we set F_{net} as the tension in the rope minus the acceleration of gravity, because Box B is moving upwards (the direction that Box A is pulling it) against gravity. The arrows on the rope in the picture should help you see this.
3. We set the force of gravity as m_{A}g or m_{B}g (versus just g) because F = m * a.
Step 3. Set acceleration from our first equation equal to acceleration from our second equation to solve for T
Step 4. Use Box A acceleration to solve for acceleration (a)
a = g – (^{T}/^{20 kg})
a = 10 m/s^{2} – (^{80 N}/_{20 kg})
a = 6 m/s^{2}Note that this acceleration is in the downward direction, which in this scenario, we have chosen to define as positive for simplicity.
Step 5. From here, we can use the displacement equation for the time it takes for the box to fall:
d = v_{i}t + ^{1}/_{2}at^{2}
12 m = 0 + ^{1}/_{2}(6 m/s^{2})t^{2}
12 m = 3t^{2}
t = 2 seconds 
Question 98 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
98. Question
98. Box A and Box B are attached by a rope over a frictionless pulley as shown in the diagram below. They are released at the same time, at the same height. Determine how long it take Box B to travel 25 meters to reach the top of the pulley. Use gravitational acceleration (g) = 10 m/s^{2}
Correct / You marked this questionStep 1. We know a few things from the get go.
 As the two boxes are attached by the rope, they will have the same acceleration (a), but in opposite directions. The downward acceleration of Box A will have the same magnitude as the upward acceleration of Box B.
 The tension in the rope is uniform throughout the rope.
Let’s look at the net force on each box to solve for acceleration (a) in each case. You’ll see momentarily how we can use the fact that acceleration is equal for both boxes in our favor.
Step 2. Compare net force Box A versus Box B
For Box A:
F_{net} = force due to gravity MINUS tension in the rope
m_{A}g – T = m_{A} (a)
a = g – (^{T}/_{m}_{A})
a = g – (^{T}/_{15 kg})For Box B:
F_{net} = tension in the rope MINUS force due to gravity
T – m_{B}g = m_{B}(a)
a = (^{T}/_{m}_{B}) – g
a = (^{T}/_{10 kg}) – gThis is probably the most difficult step in the problem. Here are a few things you must realize to get this right:
1. For Box A, we set F_{net} as force of gravity minus the tension in the rope because the Box is moving in the direction that gravity is pulling it.
2. For Box B, we set F_{net} as the tension in the rope minus the acceleration of gravity, because Box B is moving upwards (the direction that Box A is pulling it) against gravity. The arrows on the rope in the picture should help you see this.
3. We set the force of gravity as m_{A}g or m_{B}g (versus just g) because F = m * a.
Step 3. Set acceleration from our first equation equal to acceleration from our second equation to solve for T
Step 4. Use Box B acceleration to solve for acceleration (a)
a = g – (^{120 N}/_{10 kg})
a = 10 m/s^{2} – (^{120 N}/_{10 kg})
a = 2 m/s^{2}The sign is negative as it is moving in the opposite direction of box A.
Step 5. From here, we can use the displacement equation for the time it takes for the box to fall:
d = v_{i}t + ^{1}/_{2}at^{2}
25 m = 0 + ^{1}/_{2}(2 m/s^{2})t2
25 m = t^{2}
t = 5 secondsIncorrect / You marked this questionStep 1. We know a few things from the get go.
 As the two boxes are attached by the rope, they will have the same acceleration (a), but in opposite directions. The downward acceleration of Box A will have the same magnitude as the upward acceleration of Box B.
 The tension in the rope is uniform throughout the rope.
Let’s look at the net force on each box to solve for acceleration (a) in each case. You’ll see momentarily how we can use the fact that acceleration is equal for both boxes in our favor.
Step 2. Compare net force Box A versus Box B
For Box A:
F_{net} = force due to gravity MINUS tension in the rope
m_{A}g – T = m_{A} (a)
a = g – (^{T}/_{m}_{A})
a = g – (^{T}/_{15 kg})For Box B:
F_{net} = tension in the rope MINUS force due to gravity
T – m_{B}g = m_{B}(a)
a = (^{T}/_{m}_{B}) – g
a = (^{T}/_{10 kg}) – gThis is probably the most difficult step in the problem. Here are a few things you must realize to get this right:
1. For Box A, we set F_{net} as force of gravity minus the tension in the rope because the Box is moving in the direction that gravity is pulling it.
2. For Box B, we set F_{net} as the tension in the rope minus the acceleration of gravity, because Box B is moving upwards (the direction that Box A is pulling it) against gravity. The arrows on the rope in the picture should help you see this.
3. We set the force of gravity as m_{A}g or m_{B}g (versus just g) because F = m * a.
Step 3. Set acceleration from our first equation equal to acceleration from our second equation to solve for T
Step 4. Use Box B acceleration to solve for acceleration (a)
a = g – (^{120 N}/_{10 kg})
a = 10 m/s^{2} – (^{120 N}/_{10 kg})
a = 2 m/s^{2}The sign is negative as it is moving in the opposite direction of box A.
Step 5. From here, we can use the displacement equation for the time it takes for the box to fall:
d = v_{i}t + ^{1}/_{2}at^{2}
25 m = 0 + ^{1}/_{2}(2 m/s^{2})t2
25 m = t^{2}
t = 5 seconds 
Question 99 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
99. Question
99. A 1800 kg rocket launches from the ground. After 450 m, it has attained a velocity of 300 m/s. What is the magnitude of the upward force was required to propel the rocket as described? Assume gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Rocket = 1800 kg
 Distance Traveled (d) = 900 m
 Initial Velocity (v_{i}) = 0 m/s
 Final Velocity (v_{f}) = 300 m/s
Solving for: the magnitude of the upward force was required to propel the rocket
Step 2. Next, let’s work backwards. What do we need to solve for the magnitude of the upward force was required to propel the rocket?
We know that: F_{net} = ma
 There are two forces acting on the rocket. The downward force of gravity, and the upward force of the engines.
 In order to propel the rocket as described, the upward force of the engines must overcome the force of gravity.
 So, the magnitude of upward force required will be equal to the magnitude of the force of gravity on the rocket + the remaining force required to cause the velocity change.
Step 3. Solve for the force of gravity on the rocket:
F_{GRAVITY} = ma = mg
F_{GRAVITY} = 1800 kg * 10 m/s^{2}
F_{GRAVITY} = 18,000 NIf you think about it, any upward force ≤ 18,000 N would not cause the rocket to move upwards at all. You can think about this as the “threshold” force for upward movement.
Step 4. Solve for the remaining force required to cause the velocity change:
Once we apply 18,000 N of force, we have “accounted for” the force of gravity. At this point, we solve for the remaining upward force required to cause the velocity change in the same way we would if the rocket were moving horizontally with no friction.
i.e. by using the following equation:
v_{f}^{2} = v_{i}^{2} + 2ad
(300 m/s)^{2} = 0 + 2(a)(450 m)
90,000 m^{2}/s^{2} = 900 m (a)
a = 100 m/s^{2}We know that: F = ma
F = 1800 kg * 100 m/s^{2}
F = 180,000 NThus, the magnitude of the TOTAL upward force required to propel the rocket will be equal to:
(F_{GRAVITY} = 18,000 N) + (F = 180,000 N)
F_{TOTAL}= 198,000 NIncorrect / You marked this questionStep 1. First, let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Rocket = 1800 kg
 Distance Traveled (d) = 900 m
 Initial Velocity (v_{i}) = 0 m/s
 Final Velocity (v_{f}) = 300 m/s
Solving for: the magnitude of the upward force was required to propel the rocket
Step 2. Next, let’s work backwards. What do we need to solve for the magnitude of the upward force was required to propel the rocket?
We know that: F_{net} = ma
 There are two forces acting on the rocket. The downward force of gravity, and the upward force of the engines.
 In order to propel the rocket as described, the upward force of the engines must overcome the force of gravity.
 So, the magnitude of upward force required will be equal to the magnitude of the force of gravity on the rocket + the remaining force required to cause the velocity change.
Step 3. Solve for the force of gravity on the rocket:
F_{GRAVITY} = ma = mg
F_{GRAVITY} = 1800 kg * 10 m/s^{2}
F_{GRAVITY} = 18,000 NIf you think about it, any upward force ≤ 18,000 N would not cause the rocket to move upwards at all. You can think about this as the “threshold” force for upward movement.
Step 4. Solve for the remaining force required to cause the velocity change:
Once we apply 18,000 N of force, we have “accounted for” the force of gravity. At this point, we solve for the remaining upward force required to cause the velocity change in the same way we would if the rocket were moving horizontally with no friction.
i.e. by using the following equation:
v_{f}^{2} = v_{i}^{2} + 2ad
(300 m/s)^{2} = 0 + 2(a)(450 m)
90,000 m^{2}/s^{2} = 900 m (a)
a = 100 m/s^{2}We know that: F = ma
F = 1800 kg * 100 m/s^{2}
F = 180,000 NThus, the magnitude of the TOTAL upward force required to propel the rocket will be equal to:
(F_{GRAVITY} = 18,000 N) + (F = 180,000 N)
F_{TOTAL}= 198,000 N 
Question 100 of 100Physics  Units, Vectors, Kinematics, Statics, Dynamics
100. Question
100. You are riding your bike. The following graph depicts your position versus time. What is your bike doing at interval A?
Correct / You marked this questionThe slope of a position versus time graph represents the velocity of the object. A flat slope equates to zero velocity. An upward slope equates to positive velocity. A downward slope equates to negative velocity.
As you can see in interval A, the slope is becoming flatter and flatter, until it becomes completely flat a little bit after A.
Acceleration is the rate of change of velocity. In a position versus time graph, positive acceleration would equate to a slope that is increasing over time.
Over interval A, the slope is decreasing over time. Thus, the velocity is decreasing over time. This implies that your bike is decelerating over interval A.
Incorrect / You marked this questionThe slope of a position versus time graph represents the velocity of the object. A flat slope equates to zero velocity. An upward slope equates to positive velocity. A downward slope equates to negative velocity.
As you can see in interval A, the slope is becoming flatter and flatter, until it becomes completely flat a little bit after A.
Acceleration is the rate of change of velocity. In a position versus time graph, positive acceleration would equate to a slope that is increasing over time.
Over interval A, the slope is decreasing over time. Thus, the velocity is decreasing over time. This implies that your bike is decelerating over interval A.
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