AP Physics C Mechanics Quiz: Potential Energy

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Potential EnergyQuestion 1 of 20

A particle moves in a one-dimensional potential given by $U(x) = x^4 - 8x^2 + 12$ . If the particle has a total mechanical energy of $E = 4$ J, which of the following is a turning point of its motion?

x = 0

x = \sqrt{6}

x = 3

x = 2

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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Potential Energy

Practice Potential Energy 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 Potential Energy, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.

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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.

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Question 1

$x = 0$ m
$x = \sqrt{6}$ m
$x = 3$ m
$x = 2$ m (correct answer)

Explanation: Turning points occur where the kinetic energy is zero, so the total energy equals the potential energy:

E = U(x)

. We set

4 = x^4 - 8x^2 + 12

. This gives the equation

x^4 - 8x^2 + 8 = 0

. Let

y = x^2

, then

y^2 - 8y + 8 = 0

. Using the quadratic formula:

y = \frac{8 \pm \sqrt{64 - 32}}{2} = \frac{8 \pm \sqrt{32}}{2} = 4 \pm 2\sqrt{2}

. Since

y = x^2 \geq 0

, both solutions are valid:

x^2 = 4 + 2\sqrt{2} \approx 6.83

x^2 = 4 - 2\sqrt{2} = 4

. The turning points are at

x = \pm 2

m and

x = \pm\sqrt{4 + 2\sqrt{2}}

m. Of the choices given,

x = 2

m is a turning point.

Question 2

A particle of mass $m$ is located inside a uniform solid sphere of mass $M$ and radius $R$ at a distance $r$ from its center ( $r < R$ ). The gravitational force on the particle is $F_g(r) = -G\frac{Mmr}{R^3}$ . If the potential energy is defined to be zero at the center of the sphere ( $U(0)=0$ ), what is the potential energy $U(r)$ of the particle-sphere system?

$\frac{GMmr^2}{2R^3}$ (correct answer)
$-\frac{GMmr^2}{2R^3}$
$G\frac{Mmr}{R^3}$
$\frac{GMm}{R}$

Explanation: The change in potential energy is the negative of the work done by the conservative force.

U(r) - U(0) = -\int_0^r F_g(r') dr' = -\int_0^r (-G\frac{Mmr'}{R^3}) dr' = G\frac{Mm}{R^3} \int_0^r r' dr'

. Evaluating the integral gives

G\frac{Mm}{R^3} [\frac{1}{2}r'^2]_0^r = G\frac{Mm r^2}{2R^3}

. Since

U(0)=0

, we have

U(r) = \frac{GMmr^2}{2R^3}

Question 3

A satellite of mass $m$ is moved from a circular orbit of radius $R$ around a planet of mass $M$ to a new circular orbit of radius $3R$ . What is the change in the gravitational potential energy of the satellite-planet system?

An increase of $\frac{GMm}{3R}$
A decrease of $\frac{GMm}{3R}$
An increase of $\frac{2GMm}{3R}$ (correct answer)
A decrease of $\frac{2GMm}{3R}$

Explanation: The gravitational potential energy of the system is given by

U_g = -\frac{GMm}{r}

. The initial potential energy at radius

R

U_i = -\frac{GMm}{R}

. The final potential energy at radius

3R

U_f = -\frac{GMm}{3R}

. The change in potential energy is

\Delta U = U_f - U_i = -\frac{GMm}{3R} - (-\frac{GMm}{R}) = -\frac{GMm}{3R} + \frac{3GMm}{3R} = \frac{2GMm}{3R}

. Since the result is positive, it is an increase.

Question 4

A non-ideal spring exerts a restoring force given by $F(x) = -kx - \beta x^3$ , where $k$ and $\beta$ are positive constants and $x$ is the displacement from equilibrium. What is the potential energy $U(x)$ stored in the spring when it is displaced a distance $x$ , assuming $U(0) = 0$ ?

$\frac{1}{2}kx^2 - \frac{1}{4}\beta x^4$
$-\frac{1}{2}kx^2 - \frac{1}{4}\beta x^4$
$\frac{1}{2}kx^2 + \frac{1}{4}\beta x^4$ (correct answer)
$k + 3\beta x^2$

Explanation: The potential energy

U(x)

is found by integrating the negative of the force:

U(x) - U(0) = -\int_0^x F(x') dx'

. With

U(0)=0

, we have

U(x) = -\int_0^x (-kx' - \beta x'^3) dx' = \int_0^x (kx' + \beta x'^3) dx' = [\frac{1}{2}kx'^2 + \frac{1}{4}\beta x'^4]_0^x = \frac{1}{2}kx^2 + \frac{1}{4}\beta x^4

Question 5

A block of mass $m$ is attached to a vertical spring with spring constant $k$ . The spring's unstretched position is at $y=0$ . The block is held at $y=0$ and then released, after which it moves downward. What is the total potential energy of the block-spring-Earth system when the block is at a position $y < 0$ ? Let the gravitational potential energy be zero at $y=0$ .

$\frac{1}{2}ky^2 - mgy$
$mgy - \frac{1}{2}ky^2$
$\frac{1}{2}ky^2 + mgy$ (correct answer)
$-\frac{1}{2}ky^2 - mgy$

Explanation: The total potential energy is the sum of the elastic potential energy of the spring and the gravitational potential energy of the block-Earth system. The spring is stretched by a distance

|y| = -y

(since y is negative). The elastic potential energy is

U_s = \frac{1}{2}k(\text{stretch})^2 = \frac{1}{2}k(-y)^2 = \frac{1}{2}ky^2

. The gravitational potential energy relative to

y=0

U_g = mgy

. The total potential energy is

U_{total} = U_s + U_g = \frac{1}{2}ky^2 + mgy

Question 6

A block of mass $m$ is attached to a spring with spring constant $k$ . The spring is stretched a distance $x$ , storing potential energy $U$ . If the spring is cut into two equal halves and the block is attached to one of the halves, what distance must the half-spring be stretched to store the same potential energy $U$ ?

$x/\sqrt{2}$ (correct answer)
$x/2$
$x\sqrt{2}$
$2x$

Explanation: When a spring of constant

k

is cut in half, each half has a new spring constant

k' = 2k

. The original potential energy is

U = \frac{1}{2}kx^2

. We want the new potential energy

U'

to be equal to

U

. Let the new stretch distance be

x'

. Then

U' = \frac{1}{2}k'(x')^2 = \frac{1}{2}(2k)(x')^2 = k(x')^2

. Setting

U' = U

gives

k(x')^2 = \frac{1}{2}kx^2

. Solving for

x'

yields

(x')^2 = \frac{x^2}{2}

, so

x' = \frac{x}{\sqrt{2}}

Question 7

The change in gravitational potential energy for an object of mass $m$ lifted a small height $h$ near the Earth's surface (radius $R_E$ ) is well approximated by $\Delta U_g = mgh$ . The validity of this approximation, derived from the universal law of gravitation $U_g = -GM_Em/r$ , rests primarily on which assumption for $h \ll R_E$ ?

The mass of the Earth $M_E$ is much greater than the mass of the object $m$ .
The Earth can be treated as a perfect sphere of uniform density.
The gravitational force exerted by the Earth on the object is approximately constant over the height $h$ . (correct answer)
The object is lifted without any change in its kinetic energy.

Explanation: The work done by gravity is

W_g = \int F_g dr

. If the force

F_g

is approximately constant over the displacement

h

, then

W_g \approx -F_g h

. Since

W_g = -\Delta U_g

, we get

\Delta U_g \approx F_g h

. Near the surface,

F_g \approx mg

, so

\Delta U_g \approx mgh

. The core of the approximation is that the gravitational field strength

g

, and thus the force, does not change significantly over the small height

h

Question 8

A particle is moved from point P to point Q in a region where a conservative force field exists. Path 1 is a direct straight line from P to Q. Path 2 is a semicircular arc from P to Q. The work done by the conservative force is $W_1$ along Path 1 and $W_2$ along Path 2. Which of the following statements is always true?

$W_1 > W_2$ because Path 2 is longer.
$W_1 < W_2$ because Path 1 is more direct.
$W_1 = W_2$ (correct answer)
The relationship between $W_1$ and $W_2$ cannot be determined without knowing the force field.

Explanation: A defining property of a conservative force is that the work it does on a particle moving between two points is independent of the path taken. Since the force is conservative and both paths start at P and end at Q, the work done must be the same for both paths. Therefore,

W_1 = W_2

Question 9

A particle moves along the x-axis under the influence of a conservative force. Its potential energy is described by the function $U(x) = \frac{A}{x^2} - \frac{B}{x}$ , where A and B are positive constants. What is the force $F(x)$ exerted on the particle at position $x$ ?

$F(x) = \frac{2A}{x^3} - \frac{B}{x^2}$ (correct answer)
$F(x) = -\frac{2A}{x^3} + \frac{B}{x^2}$
$F(x) = -\frac{A}{x} + \frac{B}{2}$
$F(x) = \frac{A}{x} - B$

Explanation: The force is the negative gradient of the potential energy. For one dimension, this is

F(x) = -\frac{dU}{dx}

. Differentiating

U(x)

with respect to

x

gives

\frac{dU}{dx} = \frac{d}{dx}(Ax^{-2} - Bx^{-1}) = -2Ax^{-3} - (-1)Bx^{-2} = -\frac{2A}{x^3} + \frac{B}{x^2}

. Therefore,

F(x) = -(-\frac{2A}{x^3} + \frac{B}{x^2}) = \frac{2A}{x^3} - \frac{B}{x^2}

Question 10

A particle is acted upon by a one-dimensional conservative force given by $F(x) = -kx + cx^3$ , where $k$ and $c$ are positive constants. What is the change in potential energy of the particle as it moves from position $x=0$ to $x=L$ ?

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

Explanation: The change in potential energy is related to the work done by the conservative force by

\Delta U = -W_c = -\int_{x_i}^{x_f} F(x) dx

. Substituting the given force and limits:

\Delta U = -\int_0^L (-kx + cx^3) dx = -[-\frac{1}{2}kx^2 + \frac{1}{4}cx^4]_0^L = -(-\frac{1}{2}kL^2 + \frac{1}{4}cL^4) = \frac{1}{2}kL^2 - \frac{1}{4}cL^4

Question 11

The gravitational potential energy of an object of mass $m$ at a height $h$ above the ground is $U = mgh$ . This definition assumes the potential energy is zero at ground level ( $h=0$ ). If the zero point for potential energy is redefined to be at a height $h_0$ above the ground, how does this affect the calculated work done by gravity as the object moves from height $h_1$ to $h_2$ ?

The work done by gravity is unchanged. (correct answer)
The work done by gravity increases by $mgh_0$ .
The work done by gravity decreases by $mgh_0$ .
The work done by gravity depends on the new zero point $h_0$ in a more complex way.

Explanation: The work done by a conservative force, like gravity, equals the negative change in potential energy:

W_g = -\Delta U

. With the original definition,

W_g = -(mgh_2 - mgh_1) = mg(h_1 - h_2)

. With the new definition,

U' = mg(h - h_0)

, so the change is

\Delta U' = mg(h_2 - h_0) - mg(h_1 - h_0) = mgh_2 - mgh_1

. The work done is

W'_g = -\Delta U' = -(mgh_2 - mgh_1) = mg(h_1 - h_2)

. The work done is independent of the choice of the zero point for potential energy.

Question 12

A potential energy function can be defined for a force $F$ if and only if the work done by the force is independent of the path taken between any two points. For which of the following forces can a potential energy function NOT be defined?

The gravitational force between two planets.
The electrostatic force between two point charges.
The restoring force of an ideal spring.
The drag force on an object moving through a fluid. (correct answer)

Explanation: A potential energy function can only be defined for conservative forces, for which the work done is path-independent. Gravitational, electrostatic, and ideal spring forces are all conservative. The drag force, however, is a non-conservative force. It always opposes the direction of motion, so the work it does depends on the total length of the path taken, not just the start and end points. Therefore, a potential energy function cannot be defined for the drag force.

Question 13

A particle is subject to a conservative force described by $F(x) = -ax^3$ , where $a$ is a positive constant. If the potential energy is taken to be zero at $x=0$ , what is the potential energy function $U(x)$ ?

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

Explanation: The potential energy function

U(x)

is found by integrating the negative of the force:

U(x) - U(x_0) = -\int_{x_0}^x F(x') dx'

. With

U(0)=0

, we have

U(x) = -\int_0^x (-ax'^3) dx' = a \int_0^x x'^3 dx' = a [\frac{1}{4}x'^4]_0^x = \frac{1}{4}ax^4

Question 14

A particle's motion is constrained to the x-axis and is influenced by a potential energy function $U(x)$ . If the particle moves from an initial position $x_i$ to a final position $x_f$ , the work done by the conservative force associated with $U(x)$ is equal to which of the following?

$U(x_f) - U(x_i)$
$U(x_i) - U(x_f)$ (correct answer)
$U(x_f) + U(x_i)$
Zero, because the force is conservative.

Explanation: By definition, the work done by a conservative force

W_c

is equal to the negative of the change in potential energy,

\Delta U

. The change in potential energy is

\Delta U = U_{final} - U_{initial} = U(x_f) - U(x_i)

. Therefore,

W_c = -\Delta U = -(U(x_f) - U(x_i)) = U(x_i) - U(x_f)

Question 15

A particle of mass $m$ is located at a distance $x$ from the near end of a thin, uniform rod of mass $M$ and length $L$ . The particle is on the axis of the rod. Assuming the gravitational potential energy is zero at infinite separation, what is the gravitational potential energy of the particle-rod system?

$-\frac{GMm}{x+L/2}$
$-\frac{GMm}{L} \ln(1 + \frac{L}{x})$ (correct answer)
$-\frac{GMm}{x}$
$-\frac{GMm}{L}(\frac{1}{x} - \frac{1}{x+L})$

Explanation: We must integrate the potential energy contributions from all parts of the rod. Let the rod lie from

r=x

r=x+L

from the particle's perspective. A small mass element of the rod is

dM = (M/L)dr

. The potential energy of this element with the particle

m

dU = -G\frac{m dM}{r} = -G\frac{m(M/L)dr}{r}

. Integrating from

x

x+L

gives

U = \int_x^{x+L} -G\frac{Mm}{L} \frac{dr}{r} = -\frac{GMm}{L} [\ln r]_x^{x+L} = -\frac{GMm}{L}(\ln(x+L) - \ln x) = -\frac{GMm}{L} \ln(\frac{x+L}{x}) = -\frac{GMm}{L} \ln(1 + \frac{L}{x})

Question 16

A particle with total mechanical energy $E_{tot}$ oscillates in a region described by a potential energy function $U(x)$ . The kinetic energy of the particle is maximized when...

the potential energy $U(x)$ is at a local minimum. (correct answer)
the potential energy $U(x)$ is at a local maximum.
the potential energy $U(x)$ is equal to zero.
the potential energy $U(x)$ is equal to the total energy $E_{tot}$ .

Explanation: Total mechanical energy is conserved, so

E_{tot} = K + U(x)

, where

K

is the kinetic energy. Rearranging for kinetic energy gives

K = E_{tot} - U(x)

. Since

E_{tot}

is constant,

K

is maximized when the potential energy

U(x)

is minimized. This occurs at a local minimum of the potential energy function.

Question 17

A projectile is launched from the surface of a planet of mass $M$ and radius $R$ . What is the minimum total mechanical energy required for the projectile-planet system so that the projectile can escape the planet's gravitational pull? Assume the potential energy of the system is zero at infinite separation.

Zero (correct answer)
$\frac{GMm}{R}$
$-\frac{GMm}{R}$
$\frac{GMm}{2R}$

Explanation: For the projectile to escape, it must be able to reach an infinite distance from the planet with a non-negative kinetic energy (i.e., its speed must be real). At infinite separation (

r \to \infty

), the potential energy

U = -GMm/r

approaches zero. The minimum condition for escape is that the kinetic energy also approaches zero at infinite separation. By conservation of mechanical energy, the total energy

E = K + U

must be constant. If

K \to 0

and

U \to 0

r \to \infty

, the total mechanical energy of the system must be zero.

Question 18

The potential energy of a particle moving in the xy-plane is given by $U(x, y) = 3x^2y - 5y^3$ . What is the force vector $\vec{F}$ acting on the particle at the point (2, 1)?

$-12\hat{i} + 3\hat{j}$ (correct answer)
$12\hat{i} - 3\hat{j}$
$-6\hat{i} - 12\hat{j}$
$6\hat{i} + 12\hat{j}$

Explanation: The force vector is the negative gradient of the potential energy:

\vec{F} = -\nabla U = -(\frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j})

. First, find the partial derivatives:

\frac{\partial U}{\partial x} = \frac{\partial}{\partial x}(3x^2y - 5y^3) = 6xy

\frac{\partial U}{\partial y} = \frac{\partial}{\partial y}(3x^2y - 5y^3) = 3x^2 - 15y^2

. Now evaluate at the point (2, 1):

\frac{\partial U}{\partial x}|_{(2,1)} = 6(2)(1) = 12

\frac{\partial U}{\partial y}|_{(2,1)} = 3(2^2) - 15(1^2) = 12 - 15 = -3

. So,

\vec{F} = -(12\hat{i} - 3\hat{j}) = -12\hat{i} + 3\hat{j}

Question 19

Two identical masses are connected by a massless rope that passes over a massless pulley. One mass hangs vertically, while the other slides on a frictionless inclined plane that makes an angle $\theta$ with the horizontal. The system is released from rest when both masses are at the same height. After the hanging mass has descended a distance $h$ , what is the change in total potential energy of the system?

$-mgh(1 + \sin\theta)$
$-mgh(1 - \sin\theta)$ (correct answer)
$-mgh\cos\theta$
$-mgh(2 - \sin\theta)$

Explanation: The hanging mass loses potential energy

mgh

as it descends distance

h

. The mass on the incline gains potential energy as it moves up the incline. Since the rope is inextensible, when the hanging mass descends

h

, the mass on the incline moves distance

h

along the incline, gaining height

h\sin\theta

. Thus it gains potential energy

mgh\sin\theta

. The total change in potential energy is

-mgh + mgh\sin\theta = -mgh(1 - \sin\theta)

. Choice A incorrectly adds the contributions instead of subtracting. Choice C ignores the change in potential energy of the hanging mass. Choice D uses an incorrect geometric relationship.

Question 20

Two point masses $m_1$ and $m_2$ are initially separated by distance $r_i$ . They are then moved to a final separation of $r_f$ , where $r_f > r_i$ . During this process, the change in gravitational potential energy is $\Delta U$ . If the same masses are moved from separation $2r_i$ to separation $2r_f$ , what is the change in gravitational potential energy for this second process?

$\Delta U$
$\frac{\Delta U}{2}$ (correct answer)
$2\Delta U$
$\frac{\Delta U}{4}$

Explanation: The gravitational potential energy between two point masses is

U = -\frac{Gm_1m_2}{r}

. For the first process:

\Delta U = -\frac{Gm_1m_2}{r_f} - \left(-\frac{Gm_1m_2}{r_i}\right) = Gm_1m_2\left(\frac{1}{r_i} - \frac{1}{r_f}\right)

. For the second process with doubled distances:

\Delta U_2 = -\frac{Gm_1m_2}{2r_f} - \left(-\frac{Gm_1m_2}{2r_i}\right) = Gm_1m_2\left(\frac{1}{2r_i} - \frac{1}{2r_f}\right) = \frac{1}{2}Gm_1m_2\left(\frac{1}{r_i} - \frac{1}{r_f}\right) = \frac{\Delta U}{2}

. Choice A would be correct if gravitational PE were proportional to distance rather than inverse distance. Choice C incorrectly doubles the change. Choice D would result from incorrectly applying an inverse square relationship.