AP Physics C Mechanics Quiz: Work

20 questions · exam conditions

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WorkQuestion 1 of 20

A block slides at a constant velocity across a horizontal, frictionless surface. A constant vertical force of magnitude F is then applied to the block. How much work is done by this new vertical force as the block continues to travel a horizontal distance $d$ ?

Fd

-Fd

Zero

It cannot be determined without the mass of the block.

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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Work

Practice Work in AP Physics C Mechanics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Work, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$Fd$
$-Fd$
Zero (correct answer)
It cannot be determined without the mass of the block.

Explanation: Work is done only when there is a component of force in the direction of displacement. The applied force is vertical, while the block's displacement is horizontal. Since the force and displacement vectors are perpendicular to each other, the dot product of the two is zero, and no work is done by the vertical force.

Question 2

A block is attached to a horizontal spring with spring constant $k$ . The block is pulled from its equilibrium position at $x=0$ to a position $x=A$ . What is the work done by the spring on the block during this displacement?

$\frac{1}{2}kA^2$
$-\frac{1}{2}kA^2$ (correct answer)
$-kA^2$
$kA^2$

Explanation: The force exerted by the spring is

F_s = -kx

. The work done by the spring is the integral of this force over the displacement:

W_s = \int_0^A (-kx) dx = -k\left[\frac{x^2}{2}\right]_0^A = -\frac{1}{2}kA^2

. The work is negative because the spring force opposes the direction of displacement.

Question 3

An object of mass $m$ is released from rest at the top of a frictionless ramp of height $h$ and angle $\theta$ . What is the work done by the normal force on the object as it slides down the entire length of the ramp?

$mgh$
Zero (correct answer)
$-mgh$
$mg \cos\theta \cdot \frac{h}{\sin\theta}$

Explanation: The normal force exerted by the ramp on the object is, by definition, perpendicular to the surface of the ramp. The object's displacement is parallel to the surface of the ramp. Since the normal force vector is always perpendicular to the displacement vector, the dot product is zero, and the work done by the normal force is zero.

Question 4

A 5 kg block is pushed up a 30° incline of length 10 m by a constant force of 60 N parallel to the incline. The coefficient of kinetic friction between the block and the incline is 0.20. What is the net work done on the block? (Use $g=10$ m/s² and $\cos(30^\circ) \approx 0.87$ ).

600 J
350 J
263 J (correct answer)
250 J

Explanation: The net work is the sum of the work done by the applied force, gravity, and friction. Work by applied force:

W_F = (60 \text{ N})(10 \text{ m}) = 600 \text{ J}

. Work by gravity:

W_g = -mgh = -mgd\sin\theta = -(5)(10)(10)\sin(30^\circ) = -250 \text{ J}

. Normal force:

N = mg\cos\theta = (5)(10)\cos(30^\circ) \approx 43.5 \text{ N}

. Work by friction:

W_f = -f_k d = -\mu_k N d = -(0.20)(43.5)(10) = -87 \text{ J}

. Net work:

W_{net} = 600 - 250 - 87 = 263 \text{ J}

Question 5

A force is described by the vector function $\vec{F} = k(y\hat{i} - x\hat{j})$ , where $k$ is a constant. Consider two paths in the xy-plane from (0, 0) to (L, L): Path 1 is a straight line, and Path 2 goes from (0, 0) to (L, 0), then to (L, L). Let $W_1$ and $W_2$ be the work done by the force along these paths. Which statement is true?

$W_1 = W_2 = 0$
$W_1 = W_2 = -kL^2$
$W_1 = 0$ and $W_2 = -kL^2$ (correct answer)
$W_1 = -kL^2$ and $W_2 = 0$

Explanation: For Path 1, the line is

y=x

, so

dy=dx

\vec{F} = k(x\hat{i} - x\hat{j})

and

d\vec{r} = dx(\hat{i} + \hat{j})

W_1 = \int_0^L k(x-x)dx = 0

. For Path 2, the first leg is from (0,0) to (L,0), where

y=0, dy=0

W_{2a} = \int_0^L k(-x\hat{j}) \cdot (dx\hat{i}) = 0

. The second leg is from (L,0) to (L,L), where

x=L, dx=0

W_{2b} = \int_0^L k(y\hat{i} - L\hat{j}) \cdot (dy\hat{j}) = \int_0^L -kL dy = -kL^2

. Total work

W_2 = W_{2a} + W_{2b} = -kL^2

. Thus,

W_1=0

and

W_2=-kL^2

Question 6

A particle moves along a path such that its position is given by $\vec{r}(t) = (t^2 \hat{i} + 2t \hat{j})$ m. A constant force $\vec{F} = (5 \hat{i} - 3 \hat{j})$ N acts on the particle. What is the instantaneous rate at which work is being done on the particle at $t=1$ s?

4 W (correct answer)
6 W
10 W
16 W

Explanation: The instantaneous rate at which work is done is power, given by

P = \vec{F} \cdot \vec{v}

. First, find the velocity vector by taking the derivative of the position vector:

\vec{v}(t) = \frac{d\vec{r}}{dt} = (2t \hat{i} + 2 \hat{j})

m/s. At

t=1

s, the velocity is

\vec{v}(1) = (2 \hat{i} + 2 \hat{j})

m/s. Now, calculate the power:

P = (5 \hat{i} - 3 \hat{j}) \cdot (2 \hat{i} + 2 \hat{j}) = (5)(2) + (-3)(2) = 10 - 6 = 4

Question 7

A particle moves in the xy-plane under the influence of a force $\vec{F} = (2y\hat{i} + 3x\hat{j})$ N. How much work is done by this force on the particle as it moves in a straight line from the origin (0,0) to the point (2, 4) m?

16 J
20 J (correct answer)
24 J
40 J

Explanation: The path can be parameterized by

y=2x

, so

dy=2dx

. The displacement vector is

d\vec{r} = dx \hat{i} + dy \hat{j} = dx \hat{i} + 2dx \hat{j}

. Along this path, the force is

\vec{F} = (2(2x)\hat{i} + 3x\hat{j}) = (4x\hat{i} + 3x\hat{j})

. The work is

W = \int \vec{F} \cdot d\vec{r} = \int_0^2 (4x dx + (3x)(2dx)) = \int_0^2 10x dx = \left[5x^2\right]_0^2 = 5(4) = 20 \text{ J}

Question 8

In a mechanics lab, a $5.0\,\text{kg}$ block is pulled across a frictionless horizontal track by a constant force of magnitude $F=18\,\text{N}$ applied at an angle $\theta=25^\circ$ above the horizontal. The block moves a horizontal displacement of $d=4.0\,\text{m}$ . Based on the scenario described above, how much work is done by the force in pulling the object? Use $W=Fd\cos(\theta)$ .

$72\,\text{J}$
$-65\,\text{J}$
$65\,\text{J}$ (correct answer)
$30\,\text{J}$

Explanation: This question tests understanding of work in physics, specifically the calculation of work done by a force applied at an angle. Work is defined as the product of force and displacement in the direction of the force, calculated using W = Fd cos(θ). In this scenario, you are given a force of 18 N applied at an angle of 25° above the horizontal, and a horizontal displacement of 4.0 m. Choice C is correct because it accurately applies the work formula: W = (18 N)(4.0 m)cos(25°) = 72 × 0.906 = 65.2 J ≈ 65 J. Choice A incorrectly calculates 18 × 4 = 72 J without considering the angle. When teaching this concept, emphasize drawing free body diagrams to visualize the force components. Have students practice identifying which angle to use in the cosine function - it's always the angle between the force vector and displacement vector.

Question 9

A constant force $F=15\,\text{N}$ pulls a dynamics cart on a frictionless horizontal surface through a displacement $d=10\,\text{m}$ . The force is directed $\theta=90^\circ$ relative to the displacement (perpendicular). Based on the scenario described above, how much work is done by the force in pulling the object? Use $W=Fd\cos(\theta)$ .

$150\,\text{J}$
$0\,\text{J}$ (correct answer)
$15\,\text{J}$
$-150\,\text{J}$

Explanation: This question tests understanding of work when force is perpendicular to displacement. Work is defined as W = Fd cos(θ), where θ is the angle between force and displacement vectors. In this scenario, a 15 N force is applied at 90° to the 10 m displacement direction. Choice B is correct because cos(90°) = 0, giving W = (15 N)(10 m)(0) = 0 J. Choices A and D incorrectly calculate as if the force were parallel or antiparallel to displacement. This is a fundamental concept: perpendicular forces do no work because they don't contribute to motion in the displacement direction. Use examples like circular motion where centripetal force is always perpendicular to velocity, doing no work. Have students practice identifying perpendicular force-displacement pairs in various contexts.

Question 10

A $3.0\,\text{kg}$ block is pulled up a frictionless incline of angle $35^\circ$ by a constant force $F=25\,\text{N}$ applied at an angle of $15^\circ$ above the incline surface. The block moves $d=2.0\,\text{m}$ along the incline. The angle between the applied force and the displacement is $\theta=15^\circ$ . Based on the scenario described above, calculate the work done in moving the object up the incline using $W=Fd\cos(\theta)$ .

$50\,\text{J}$
$13\,\text{J}$
$48\,\text{J}$ (correct answer)
$-48\,\text{J}$

Explanation: This question tests understanding of work when force is applied at an angle above an inclined surface. Work is calculated as W = Fd cos(θ), where θ is the angle between force and displacement. A 25 N force is applied 15° above the incline surface, moving the block 2.0 m along the incline, so the angle between force and displacement is 15°. Choice C is correct: W = (25 N)(2.0 m)cos(15°) = 50 × 0.966 = 48.3 J ≈ 48 J. Choice D incorrectly assigns a negative value, which would occur if force opposed motion. When teaching incline problems with angled forces, carefully distinguish between angles relative to horizontal, relative to incline, and between force-displacement vectors. Use clear diagrams showing all relevant angles to prevent confusion.

Question 11

An elevator motor lifts a $650\,\text{kg}$ load vertically upward through a height $h=12\,\text{m}$ at constant speed. Take $g=9.8\,\text{m/s}^2$ . The motor's cable tension is upward and approximately equal to the weight, so the force is parallel to the displacement ( $\theta=0^\circ$ ). Based on the scenario described above, determine the work required to lift the mass to the specified height using $W=Fd\cos(\theta)$ .

$-7.6\times10^4\,\text{J}$
$7.6\times10^4\,\text{J}$ (correct answer)
$6.4\times10^3\,\text{J}$
$7.8\times10^4\,\text{J}$

Explanation: This question tests understanding of work done against gravity when lifting an object vertically. Work is the product of force and displacement in the direction of the force, W = Fd cos(θ). In this scenario, the motor must apply an upward force equal to the weight (mg = 650 kg × 9.8 m/s² = 6370 N) to lift at constant speed, moving the load 12 m upward. Choice B is correct because the force and displacement are both upward (θ = 0°), so W = (6370 N)(12 m)(1) = 76,440 J ≈ 7.6×10⁴ J. Choice A incorrectly assigns a negative sign, which would apply if the force opposed the motion. When teaching, emphasize that work is positive when force and displacement are in the same direction. Have students identify force direction relative to motion to determine the sign of work.

Question 12

A $4.0\,\text{kg}$ crate is pulled along a frictionless floor by a constant force $F=50\,\text{N}$ directed $60^\circ$ above the horizontal. The crate moves $d=3.0\,\text{m}$ horizontally. Based on the scenario described above, how much work is done by the force in pulling the object? Use $W=Fd\cos(\theta)$ with $\theta=60^\circ$ .

$150\,\text{J}$
$75\,\text{J}$ (correct answer)
$130\,\text{J}$
$-75\,\text{J}$

Explanation: This question tests understanding of work when force is applied at an angle to the displacement. Work equals the product of force and displacement in the direction of motion, W = Fd cos(θ). In this scenario, a 50 N force is applied at 60° above the horizontal while the crate moves 3.0 m horizontally. Choice B is correct because it properly applies the formula: W = (50 N)(3.0 m)cos(60°) = 150 × 0.5 = 75 J. Choice A incorrectly calculates 50 × 3 = 150 J without considering the angle. When teaching this concept, use component analysis to show that only the horizontal component of force (F cos 60°) contributes to work. Practice problems should vary the angle to reinforce that work depends on the force component parallel to displacement.

Question 13

A particle moves along the x-axis under the influence of a force given by $F(x) = kx^3$ , where $k$ is a constant. What is the work done by this force as the particle moves from $x = 0$ to $x = L$ ?

$kL^3$
$3kL^2$
$\frac{1}{3}kL^3$
$\frac{1}{4}kL^4$ (correct answer)

Explanation: Work done by a variable force is the integral of the force with respect to displacement.

W = \int_{x_i}^{x_f} F(x) dx

. In this case,

W = \int_{0}^{L} kx^3 dx = k \left[ \frac{x^4}{4} \right]_{0}^{L} = k \left( \frac{L^4}{4} - 0 \right) = \frac{1}{4}kL^4

Question 14

A force acting on a particle moving along the x-axis is given by a function $F(x)$ . The work done by the force as the particle moves from $x=x_1$ to $x=x_2$ is determined by which of the following?

The slope of the $F(x)$ versus $x$ graph at the midpoint between $x_1$ and $x_2$ .
The average value of the force multiplied by the final position $x_2$ .
The change in the force, $F(x_2) - F(x_1)$ , multiplied by the displacement.
The area under the $F(x)$ versus $x$ graph between $x_1$ and $x_2$ . (correct answer)

Explanation: The work done by a variable force is given by the integral

W = \int_{x_1}^{x_2} F(x) dx

. The definite integral of a function between two points represents the area under the curve of the function's graph between those two points. Therefore, the work done is the area under the force versus displacement graph.

Question 15

An object is moved from point P to point Q along two different paths, Path 1 and Path 2. The gravitational force and a kinetic friction force are the only forces doing work on the object. Path 2 is longer than Path 1. Let $W_{g1}$ and $W_{f1}$ be the work done by gravity and friction along Path 1, and $W_{g2}$ and $W_{f2}$ be the work done along Path 2. Which statement is correct?

$W_{g1} > W_{g2}$ and $|W_{f1}| < |W_{f2}|$
$W_{g1} = W_{g2}$ and $|W_{f1}| < |W_{f2}|$ (correct answer)
$W_{g1} = W_{g2}$ and $W_{f1} = W_{f2}$
$W_{g1} < W_{g2}$ and $|W_{f1}| = |W_{f2}|$

Explanation: Gravity is a conservative force, so the work it does depends only on the change in vertical position, not on the path taken. Since the start and end points (P and Q) are the same, the change in vertical position is the same for both paths, thus

W_{g1} = W_{g2}

. Friction is a nonconservative force, and the work it does is path-dependent. The work done by friction is

W_f = -f_k \times (\text{path length})

. Since Path 2 is longer, the magnitude of the negative work done by friction will be greater along Path 2, meaning

|W_{f1}| < |W_{f2}|

Question 16

A student holds a 10 kg box at a constant height of 1.5 m above the floor for 30 seconds. The student claims that the work they do on the box is $(10 \text{ kg})(10 \text{ m/s}^2)(1.5 \text{ m}) = 150 \text{ J}$ . Why is this claim incorrect? (Use $g=10$ m/s²).

The work done is zero because the displacement of the box is zero. (correct answer)
The work done is negative because the force is upward but time passes.
The work done is actually power, and should be divided by 30 s.
The work done is zero because the net force on the box is zero.

Explanation: Work is defined as the product of the component of a force in the direction of displacement and the magnitude of the displacement (

W = \vec{F} \cdot \vec{d}

). Although the student is applying an upward force to counteract gravity, the box is not moving; its displacement is zero. Therefore, the work done by the student on the box during this time is zero.

Question 17

A force $\vec{F} = (2x\hat{i} + 3y^2\hat{j} - 4z^3\hat{k})$ N acts on a particle. What is the work done by this force as the particle moves from the origin (0, 0, 0) to the point (1, 1, 1) m? The force is conservative.

0 J
1 J (correct answer)
2 J
3 J

Explanation: Since the force is conservative, the work done is independent of the path and is equal to the negative change in potential energy,

W = -\Delta U

. The potential energy function is found by integrating the components of the force:

U(x,y,z) = -\int F_x dx - \int F_y dy - \int F_z dz = -x^2 - y^3 + z^4

. The work done is

W = -[U(1,1,1) - U(0,0,0)] = -[(-1^2 - 1^3 + 1^4) - (0)] = -[-1 - 1 + 1] = -[-1] = 1 \text{ J}

Question 18

A block on a horizontal surface is attached to a spring and oscillates back and forth. Consider the work done by the spring on the block as it moves from its maximum positive displacement $+A$ to the equilibrium position $x=0$ . Which of the following statements is correct?

The work done is positive, and the block's speed increases. (correct answer)
The work done is negative, and the block's speed decreases.
The work done is positive, but the block's speed decreases.
The work done is zero, and the block's speed is constant.

Explanation: As the block moves from

x=+A

x=0

, its displacement is in the negative x-direction. The spring force is a restoring force, so it is directed towards equilibrium, also in the negative x-direction. Since the force and displacement are in the same direction, the work done by the spring is positive. According to the work-energy theorem, positive net work results in an increase in kinetic energy, so the block's speed increases.

Question 19

A spring with spring constant $k$ is compressed by distance $x_0$ from its natural length. A mass $m$ is placed against the compressed spring and released. The mass slides along a frictionless surface and then up a frictionless inclined plane of angle $\theta$ . What is the work done by gravity on the mass when it has traveled a distance $s$ up the incline?

$-mgs$
$-mgs\sin\theta$ (correct answer)
$mgs\sin\theta$
$-mgs\cos\theta$

Explanation: Work done by gravity is

W = \vec{F}_g \cdot \vec{d}

. The gravitational force is

mg

downward, and the displacement is

s

up the incline. The component of displacement in the direction of gravity is

-s\sin\theta

(negative because it's opposite to gravity). Therefore,

W = mg \cdot (-s\sin\theta) = -mgs\sin\theta

. Choice A ignores the incline geometry. Choice C has the wrong sign. Choice D uses cosine instead of sine.

Question 20

A particle moves in a circle of radius $R$ under the influence of a central force $\vec{F} = -kr^2\hat{r}$ , where $k > 0$ and $\hat{r}$ is the radial unit vector. As the particle completes one full revolution, what is the total work done by this force?

$-2\pi kR^3$
$0$ (correct answer)
$-kR^3$
$\pi kR^3$

Explanation: For circular motion at constant radius

R

, the displacement

d\vec{r}

is always tangential to the circle, while the central force

\vec{F} = -kR^2\hat{r}

is always radial. Since

\vec{F} \perp d\vec{r}

at every point,

\vec{F} \cdot d\vec{r} = 0

, so the total work is zero. Choice A incorrectly assumes work is done despite perpendicularity. Choice C represents work for partial motion. Choice D has wrong sign and assumes non-zero work.