AP Physics C Mechanics Quiz: Conservation Of Energy

20 questions · exam conditions

0:00

Conservation Of EnergyQuestion 1 of 20

A roller coaster car of mass $m$ travels along a frictionless track with several hills and valleys. Which of the following statements about the work done by the normal force, $W_N$ , exerted by the track on the car is correct?

W_N

is always positive because the normal force must support the car's weight.

W_N

is always zero because the normal force is always perpendicular to the car's velocity.

W_N

is positive on uphill sections and negative on downhill sections to assist the motion.

W_N

depends on the car's speed because a higher speed requires a larger normal force in curves.

← Back to quizzes

AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Conservation Of Energy

Practice Conservation Of 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 Conservation Of Energy, 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

$W_N$ is always positive because the normal force must support the car's weight.
$W_N$ is always zero because the normal force is always perpendicular to the car's velocity. (correct answer)
$W_N$ is positive on uphill sections and negative on downhill sections to assist the motion.
$W_N$ depends on the car's speed because a higher speed requires a larger normal force in curves.

Explanation: Work is defined by the dot product of force and displacement,

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

. The normal force is, by definition, always perpendicular to the surface of the track. The instantaneous displacement (and thus velocity) of the car is always tangent to the track. Since the normal force and displacement vectors are always perpendicular, their dot product is zero, and the normal force does no work.

Question 2

A small block is released from rest at height $H$ on a frictionless track that includes a circular loop of radius $R$ . What is the minimum height $H$ from which the block must be released to remain in contact with the track at the top of the loop?

$R$
$2R$
$2.5R$ (correct answer)
$3R$

Explanation: To remain in contact at the top of the loop (height

2R

), the centripetal force must at least equal the gravitational force, so

mg = mv_{top}^2/R

, which gives the minimum speed

v_{top}^2 = gR

. By conservation of energy from the start to the top of the loop:

mgH = mg(2R) + \frac{1}{2}mv_{top}^2

. Substituting

v_{top}^2

mgH = 2mgR + \frac{1}{2}mgR = 2.5mgR

. Thus,

H = 2.5R

Question 3

A ball is thrown vertically upward from the ground with an initial kinetic energy $K_0$ . It reaches a maximum height where its gravitational potential energy is $U_{max}$ . During its flight, it is subject to a non-zero air resistance force. Which of the following correctly relates $K_0$ and $U_{max}$ ?

$K_0 = U_{max}$
$K_0 > U_{max}$ (correct answer)
$K_0 < U_{max}$
The relationship depends on the trajectory of the ball.

Explanation: Air resistance is a non-conservative force that does negative work on the ball as it moves upward, dissipating mechanical energy. The change in mechanical energy is equal to the work done by non-conservative forces:

W_{nc} = E_f - E_i

. Here,

E_i = K_0

and

E_f = U_{max}

. Since

W_{air} < 0

, we have

U_{max} - K_0 < 0

, which implies

K_0 > U_{max}

Question 4

A bullet of mass $m$ and speed $v$ strikes a block of mass $M$ ( $M > m$ ), which is suspended by a light string of length $L$ . The bullet embeds itself in the block. Which expression represents the maximum height $h$ to which the block-bullet system swings?

$\frac{v^2}{2g}$
$\frac{m^2 v^2}{2g(M+m)^2}$ (correct answer)
$\frac{m v^2}{2g M}$
$\frac{(M+m)v^2}{2g m^2}$

Explanation: First, use conservation of momentum for the perfectly inelastic collision:

mv = (M+m)V

, where

V

is the speed just after impact. So

V = \frac{mv}{M+m}

. Then, use conservation of mechanical energy for the swing:

\frac{1}{2}(M+m)V^2 = (M+m)gh

. Solving for

h

gives

h = \frac{V^2}{2g}

. Substituting the expression for

V

gives

h = \frac{1}{2g} (\frac{mv}{M+m})^2 = \frac{m^2 v^2}{2g(M+m)^2}

Question 5

A mass $m$ on a frictionless horizontal surface is attached to a spring of constant $k$ . The mass is displaced by an amplitude $A$ from equilibrium and released from rest. At what displacement $x$ from equilibrium is the kinetic energy of the mass equal to its potential energy?

$x = A/\sqrt{2}$ (correct answer)
$x = A/2$
$x = A/\sqrt{3}$
$x = A/3$

Explanation: The total energy is constant and equal to the initial potential energy:

E_{total} = \frac{1}{2}kA^2

. We want the position

x

where kinetic energy

K

equals potential energy

U

. At this point,

E_{total} = K + U = U + U = 2U

. So,

\frac{1}{2}kA^2 = 2(\frac{1}{2}kx^2) = kx^2

. Solving for

x

gives

x^2 = A^2/2

, so

x = A/\sqrt{2}

Question 6

A block of mass $m$ slides from rest down a frictionless incline of height $h$ . It then moves onto a rough horizontal surface with a coefficient of kinetic friction $\mu_k$ . How far does the block slide on the horizontal surface before coming to rest?

$\frac{h}{\mu_k}$ (correct answer)
$\frac{gh}{\mu_k}$
$\mu_k h$
$\frac{2gh}{\mu_k}$

Explanation: On the incline, conservation of energy gives the block's kinetic energy at the bottom as

K = mgh

. On the horizontal surface, the work done by friction,

W_f = -f_k d = -\mu_k mgd

, must equal the change in kinetic energy, which is

0 - K = -mgh

. Setting

-\mu_k mgd = -mgh

and solving for

d

gives

d = h/\mu_k

Question 7

A block of mass $m$ attached to a horizontal spring of constant $k$ is displaced from its equilibrium position by a distance $A$ and released from rest. What is the speed of the block when it is at a position $x = A/2$ ?

$\sqrt{\frac{3kA^2}{4m}}$ (correct answer)
$\sqrt{\frac{kA^2}{2m}}$
$\sqrt{\frac{kA^2}{4m}}$
$\sqrt{\frac{kA}{2m}}$

Explanation: The total mechanical energy of the system, set at the moment of release, is

E = \frac{1}{2}kA^2

. At any position

x

, the energy is

E = \frac{1}{2}kx^2 + \frac{1}{2}mv^2

. By conservation of energy,

\frac{1}{2}kA^2 = \frac{1}{2}k(A/2)^2 + \frac{1}{2}mv^2

. This simplifies to

\frac{1}{2}kA^2 = \frac{1}{8}kA^2 + \frac{1}{2}mv^2

, which gives

\frac{3}{8}kA^2 = \frac{1}{2}mv^2

. Solving for

v

yields

v = \sqrt{\frac{3kA^2}{4m}}

Question 8

An object of mass 2 kg, initially at rest at the origin, is acted upon by a force given by $F(x) = 6x^2$ N, where $x$ is in meters. What is the kinetic energy of the object when it reaches $x = 2$ m?

16 J (correct answer)
8 J
24 J
32 J

Explanation: According to the work-energy theorem, the work done on the object equals its change in kinetic energy. Since it starts from rest, its final kinetic energy is equal to the work done. The work is calculated by integrating the force over the displacement:

W = \int_{0}^{2} F(x) dx = \int_{0}^{2} 6x^2 dx = [2x^3]_0^2 = 2(2)^3 - 0 = 16

Question 9

A block of mass $m$ is dropped from a height $h$ above the top of a vertical spring with spring constant $k$ . The block sticks to the spring and compresses it. Which equation must be solved to find the maximum compression $x$ of the spring?

$mgh = \frac{1}{2}kx^2$
$mg(h+x) = \frac{1}{2}kx^2$ (correct answer)
$mgh = \frac{1}{2}k(h+x)^2$
$mgx = \frac{1}{2}kx^2 - mgh$

Explanation: We apply conservation of mechanical energy between the initial point (height

h

above the spring) and the final point (maximum compression

x

). The total vertical distance the block falls is

h+x

. This loss in gravitational potential energy is converted into elastic potential energy in the spring. Initial energy (relative to max compression point) is

E_i = mg(h+x)

. Final energy is

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

. Setting

E_i = E_f

gives

mg(h+x) = \frac{1}{2}kx^2

Question 10

A car of mass $m$ travels at a constant speed $v$ up an incline with an angle $\theta$ to the horizontal. The car is subject to a constant frictional force $f$ . What is the power supplied by the engine?

$mgv \sin\theta$
$(mg \sin\theta + f)v$ (correct answer)
$(mg \cos\theta + f)v$
$(f - mg \sin\theta)v$

Explanation: Since the car moves at a constant speed, the net force on it is zero. The engine must provide a force

F_{engine}

that balances both the component of gravity down the incline (

mg \sin\theta

) and the frictional force (

f

). So,

F_{engine} = mg \sin\theta + f

. Power is given by

P = F_{engine} \cdot v

. Since the force is in the direction of velocity,

P = (mg \sin\theta + f)v

Question 11

A block of mass $m$ is released from rest at a height $h$ . In Case 1, the block falls vertically. In Case 2, the block slides down a frictionless curved ramp from the same height. Air resistance is negligible. Let $v_1$ and $v_2$ be the speeds of the block just before it reaches ground level in Case 1 and Case 2, respectively. Which of the following is true?

$v_1 > v_2$
$v_2 > v_1$
$v_1 = v_2$ (correct answer)
The relationship depends on the specific shape of the ramp.

Explanation: In both cases, the only force doing work is gravity, which is a conservative force. Therefore, the total mechanical energy of the block-Earth system is conserved. The initial energy is entirely potential,

E_i = mgh

. The final energy is entirely kinetic,

E_f = \frac{1}{2}mv^2

. Setting

E_i = E_f

gives

mgh = \frac{1}{2}mv^2

, so

v = \sqrt{2gh}

in both cases. The path does not affect the final speed if there are no non-conservative forces.

Question 12

A particle of mass $m$ moves along the x-axis under the influence of a potential energy function $U(x) = \frac{1}{4}x^4 - 2x^2$ . If the total mechanical energy of the particle is $E = -3$ J, what are the turning points of its motion?

$x = \pm \sqrt{2}$ and $x = \pm \sqrt{6}$ (correct answer)
$x = \pm 2$
$x = \pm \sqrt{2}$ only
The particle cannot have this energy.

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

E = U(x)

. We solve

-3 = \frac{1}{4}x^4 - 2x^2

. Let

y = x^2

. The equation becomes

y^2 - 8y + 12 = 0

, which factors to

(y-2)(y-6) = 0

. The solutions are

y=2

and

y=6

. Since

y=x^2

, we have

x^2 = 2

and

x^2 = 6

, which gives four turning points:

x = \pm \sqrt{2}

and

x = \pm \sqrt{6}

Question 13

The escape velocity from the surface of a planet of mass $M$ and radius $R$ is the minimum initial speed an object must have to escape the planet's gravitational pull completely. Which statement correctly describes the energy condition for an object launched with escape velocity?

Its initial kinetic energy is equal in magnitude to its initial gravitational potential energy.
Its total mechanical energy is zero. (correct answer)
Its total mechanical energy is positive, allowing it to overcome the potential barrier.
Its kinetic energy becomes zero at a finite but very large distance from the planet.

Explanation: To 'escape completely' means to reach an infinite distance (

r \to \infty

) with zero kinetic energy. The gravitational potential energy,

U = -GMm/r

, is defined to be zero at

r = \infty

. Thus, the total mechanical energy at infinity is zero. By conservation of energy, the total mechanical energy at the surface must also be zero.

E = K + U = \frac{1}{2}mv_{esc}^2 - \frac{GMm}{R} = 0

. This makes B correct and A also correct, but B is the more fundamental energy condition.

Question 14

Two blocks of masses $m_1$ and $m_2$ ( $m_1 > m_2$ ) are connected by a light string that passes over a frictionless, massless pulley. The system is released from rest. Using conservation of energy, what is the speed of the blocks after $m_1$ has descended a distance $h$ ?

$\sqrt{2gh}$
$\sqrt{2gh \frac{m_1}{m_1+m_2}}$
$\sqrt{2gh \frac{m_1-m_2}{m_1+m_2}}$ (correct answer)
$\sqrt{gh \frac{m_1-m_2}{m_1+m_2}}$

Explanation: For the system, the loss in potential energy is converted to kinetic energy. The change in potential energy is

\Delta U = -m_1gh + m_2gh = -(m_1-m_2)gh

. The gain in kinetic energy is

\Delta K = \frac{1}{2}m_1v^2 + \frac{1}{2}m_2v^2 = \frac{1}{2}(m_1+m_2)v^2

. By conservation of energy,

\Delta K + \Delta U = 0

, so

\frac{1}{2}(m_1+m_2)v^2 = (m_1-m_2)gh

. Solving for

v

gives the result.

Question 15

A small block is released from rest at the rim of a large, frictionless, hemispherical bowl of radius $R$ . What is the work done by the gravitational force on the block as it slides from the rim to the bottom of the bowl?

$mgR$ (correct answer)
$\frac{1}{2}mgR$
$0$
$-mgR$

Explanation: The work done by the conservative gravitational force is equal to the negative change in gravitational potential energy,

W_g = -\Delta U_g

. Let the bottom of the bowl be the zero reference level for potential energy (

h=0

). The rim is at a height

h=R

. So,

\Delta U_g = U_f - U_i = 0 - mgR = -mgR

. Therefore,

W_g = -(-mgR) = mgR

. Alternatively, work done by gravity is the force (

mg

) times the vertical displacement (

R

Question 16

A pendulum consists of a mass $m$ attached to a string of length $L$ . The mass is pulled to one side so the string makes an angle $\theta_0$ with the vertical, then released from rest. At the lowest point of its swing, the string breaks. Assuming the pendulum bob was initially displaced by a small angle ( $\sin\theta_0 \approx \theta_0$ , $\cos\theta_0 \approx 1 - \frac{\theta_0^2}{2}$ ), what is the horizontal distance the mass travels before hitting the ground, which is a distance $h$ below the lowest point of the swing?

$\theta_0\sqrt{\frac{2hL}{g}}$ (correct answer)
$\theta_0\sqrt{\frac{hL}{g}}$
$\theta_0L\sqrt{\frac{2h}{g}}$
$\frac{\theta_0L}{2}\sqrt{\frac{2h}{g}}$

Explanation: Using conservation of energy from release to bottom:

mg(L - L\cos\theta_0) = \frac{1}{2}mv^2

. With small angle approximation:

mgL(1 - (1 - \frac{\theta_0^2}{2})) = \frac{1}{2}mv^2

, so

v = \theta_0\sqrt{gL}

. After the string breaks, projectile motion gives

t = \sqrt{\frac{2h}{g}}

and horizontal distance

x = vt = \theta_0\sqrt{gL} \cdot \sqrt{\frac{2h}{g}} = \theta_0\sqrt{2hL}

. Choice B is missing the factor of

\sqrt{2}

. Choice C incorrectly includes an extra factor of

L

. Choice D incorrectly includes both an extra

L

factor and a factor of

\frac{1}{2}

Question 17

A block of mass $m$ is released from rest at the top of a frictionless inclined plane of height $h$ and angle $\theta$ . At the bottom of the incline, it enters a horizontal surface with coefficient of kinetic friction $\mu_k$ and slides a distance $d$ before coming to rest. If the same block is instead released from rest at height $2h$ on the same incline, what distance will it slide on the horizontal surface before stopping?

$2d$ (correct answer)
$4d$
$\sqrt{2}d$
$2\sqrt{2}d$

Explanation: Using conservation of energy: Initial potential energy

mgh

equals work done by friction

\mu_k mg d

, so

mgh = \mu_k mg d

, giving

d = \frac{h}{\mu_k}

. When released from height

2h

, the initial potential energy is

2mgh

, so

2mgh = \mu_k mg d'

, giving

d' = \frac{2h}{\mu_k} = 2d

. Choice B incorrectly assumes kinetic energy scales as velocity squared without considering the linear relationship between potential energy and work. Choice C incorrectly applies

\sqrt{2}

scaling from kinematics. Choice D combines both errors from B and C.

Question 18

A spring with spring constant $k$ is compressed by distance $x_0$ from its natural length. A block of mass $m$ is placed against the compressed spring and released. The block slides up a frictionless incline of angle $30°$ and comes to rest after traveling a distance $L$ along the incline. If the same spring is compressed by $2x_0$ and the experiment repeated with a block of mass $2m$ , how far up the incline will this block travel?

$\frac{L}{2}$
$L$
$2L$ (correct answer)
$4L$

Explanation: Initially:

\frac{1}{2}kx_0^2 = mgL\sin(30°) = \frac{mgL}{2}

, so

L = \frac{kx_0^2}{mg}

. In the second case:

\frac{1}{2}k(2x_0)^2 = 2kx_0^2

of elastic potential energy. Setting this equal to gravitational potential energy:

2kx_0^2 = (2m)gL'\sin(30°) = mgL'

. Therefore

L' = \frac{2kx_0^2}{mg} = 2L

. Choice A results from incorrectly thinking both the doubled compression and doubled mass reduce the distance. Choice B incorrectly assumes the effects of doubled compression and doubled mass cancel exactly. Choice D incorrectly applies only the compression factor without accounting for increased mass.

Question 19

A uniform rod of length $L$ and mass $M$ is pivoted at one end and released from rest in a horizontal position. Using conservation of energy, what is the angular velocity of the rod when it reaches the vertical position?

$\sqrt{\frac{6g}{L}}$
$\sqrt{\frac{3g}{2L}}$
$\sqrt{\frac{g}{L}}$
$\sqrt{\frac{3g}{L}}$ (correct answer)

Explanation: When you encounter a rotating rigid body problem involving energy conservation, you're dealing with the conversion between gravitational potential energy and rotational kinetic energy. The key insight is identifying the proper moment of inertia and the change in height of the center of mass. For a uniform rod pivoting about one end, the moment of inertia is

I = \frac{1}{3}ML^2

. When the rod falls from horizontal to vertical, its center of mass (located at

L/2

from the pivot) drops by a distance

\Delta h = \frac{L}{2}

. Using conservation of energy: Initial potential energy equals final rotational kinetic energy.

Mg \cdot \frac{L}{2} = \frac{1}{2}I\omega^2

Substituting the moment of inertia:

\frac{MgL}{2} = \frac{1}{2} \cdot \frac{1}{3}ML^2 \cdot \omega^2

Simplifying:

\frac{MgL}{2} = \frac{ML^2\omega^2}{6}

Solving for

\omega

\omega^2 = \frac{3g}{L}

\omega = \sqrt{\frac{3g}{L}}

This confirms answer D is correct. Answer A (

\sqrt{\frac{6g}{L}}

) likely comes from incorrectly using

I = \frac{1}{2}ML^2

(disk formula) instead of the rod formula. Answer B (

\sqrt{\frac{3g}{2L}}

) results from using the wrong height change or moment of inertia. Answer C (

\sqrt{\frac{g}{L}}

) resembles simple pendulum frequency, which doesn't apply here. Remember: Always verify you're using the correct moment of inertia formula for the specific geometry and pivot location. The AP Physics C reference sheet provides these formulas.

Question 20

A particle moves in a conservative force field where the potential energy is given by $U(x) = ax^2 - bx^3$ , where $a$ and $b$ are positive constants. The particle starts from rest at $x = 0$ . What is the maximum value of $x$ the particle can reach if its total mechanical energy is $E$ ?

The positive root of $ax^2 - bx^3 - E = 0$
$\sqrt{\frac{E}{a}}$
$\sqrt[3]{\frac{E}{b}}$
The positive root of $ax^2 - bx^3 = E$ (correct answer)

Explanation: When you encounter problems involving conservative forces and potential energy, think about energy conservation. The key insight is that a particle can only reach positions where its kinetic energy remains non-negative, since

KE = \frac{1}{2}mv^2 \geq 0

. Since the particle starts from rest at

x = 0

, its initial kinetic energy is zero and its initial potential energy is

U(0) = 0

. Therefore, the total mechanical energy is

E = KE_0 + U_0 = 0 + 0 = E

, which means this energy

E

must be supplied to the system initially. At any position

x

, energy conservation gives us:

E = KE + U(x) = \frac{1}{2}mv^2 + ax^2 - bx^3

. For the particle to physically reach position

x

, we need

KE \geq 0

, which means

E - U(x) \geq 0

, or

E \geq ax^2 - bx^3

. The maximum value of

x

occurs when the kinetic energy just reaches zero, meaning all energy is potential energy. At this turning point:

E = U(x) = ax^2 - bx^3

, which gives us

ax^2 - bx^3 = E

. Choice A incorrectly subtracts

E

instead of setting the potential energy equal to

E

. Choice B assumes the cubic term is negligible and only considers the quadratic term

ax^2 = E

. Choice C makes the opposite error, ignoring the quadratic term and setting

bx^3 = E

. Choice D correctly represents the condition where all mechanical energy equals potential energy at the turning point. Remember: turning points in conservative force problems occur where kinetic energy equals zero, so total energy equals potential energy.