AP Physics 1 Quiz: Conservation Of Energy

20 questions · exam conditions

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Conservation Of EnergyQuestion 1 of 20

A spring ( $k$ ) launches a block ( $m$ ) on a frictionless horizontal surface from compression $x$ . What is the block's speed when the spring reaches equilibrium?

v=\sqrt{\dfrac{kx}{m}}

v=\sqrt{\dfrac{kx^2}{m}}

v=x\sqrt{\dfrac{k}{m}}

v=\dfrac{kx^2}{2m}

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AP Physics 1 Quiz

AP Physics 1 Quiz: Conservation Of Energy

Practice Conservation Of Energy in AP Physics 1 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 1.

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

A spring ( $k$ ) launches a block ( $m$ ) on a frictionless horizontal surface from compression $x$ . What is the block's speed when the spring reaches equilibrium?

$v=\sqrt{\dfrac{kx}{m}}$
$v=\sqrt{\dfrac{kx^2}{m}}$
$v=x\sqrt{\dfrac{k}{m}}$ (correct answer)
$v=\dfrac{kx^2}{2m}$

Explanation: This problem involves energy conservation with a spring launching a block on a frictionless surface. Initially, all energy is stored as elastic potential energy in the compressed spring (

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

), and finally all energy is kinetic when the spring reaches equilibrium. The energy conservation equation is:

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

, where

x

is the initial compression and

v

is the final speed. Solving for

v

: cancel the

\frac{1}{2}

to get

kx^2 = mv^2

, divide by

m

to get

\frac{kx^2}{m} = v^2

, then take the square root:

v = x\sqrt{\frac{k}{m}}

. Choice B (

v = \sqrt{\frac{kx^2}{m}}

) is incorrect because it places

x^2

inside the square root instead of factoring out

x

. When working with spring energy problems, remember that elastic potential energy is proportional to

x^2

, so one factor of

x

comes outside the square root.

Question 2

A skier of mass $m$ starts from rest at height $h$ and slides down a slope. Air resistance is present and does negative work of magnitude $W_{\text{air}}$ during the descent. The skier reaches the bottom with speed $v$ . Which statement correctly relates the energies?

$mgh = \tfrac12 mv^2$ because gravitational potential converts entirely to kinetic energy
$mgh - W_{\text{air}} = \tfrac12 mv^2$ (correct answer)
$mgh = \tfrac12 mv^2 - W_{\text{air}}$
$mgh = \tfrac12 m\vec v^2 + \vec W_{\text{air}}$

Explanation: This question explores conservation of energy with air resistance as a non-conservative force. The skier's initial gravitational potential energy is mgh, and air resistance does negative work -W_air during descent. The energy equation is initial potential plus work by air resistance equals final kinetic energy. Thus, mgh - W_air = ½ m v² accurately describes the situation. Choice C is incorrect as it subtracts W_air from the kinetic energy, which would imply air resistance increases potential energy. Remember to treat non-conservative work as reducing the mechanical energy available for conversion to kinetic energy in such scenarios.

Question 3

Two identical blocks start from rest at the same height $h$ on two different ramps. Ramp 1 is frictionless; Ramp 2 has kinetic friction that does negative work of magnitude $W_f$ on the block. Air resistance is negligible. Both reach the bottom. Which comparison of their bottom speeds is correct?

$v_2>v_1$ because friction adds thermal energy to increase speed
$v_2=v_1$ because both lose the same gravitational potential energy $mgh$
$v_2<v_1$ because friction reduces the kinetic energy at the bottom (correct answer)
$\vec v_2=\vec v_1-\vec W_f$

Explanation: This question evaluates the effect of friction on conservation of energy in ramp systems. For the frictionless ramp, all gravitational potential mgh converts to kinetic energy ½ m v₁². On the frictional ramp, friction dissipates energy as heat, reducing the final kinetic energy, so v₂ < v₁. This comparison holds because friction does negative work, leading to less speed. Choice A is incorrect as it claims friction adds thermal energy to increase speed, which contradicts energy dissipation. When comparing systems, calculate or reason about energy losses to predict qualitative outcomes like speed differences.

Question 4

A spring with constant $k$ is compressed by distance $x$ and launches a cart of mass $m$ along a horizontal track. The track has kinetic friction coefficient $\mu_k$ over the first distance $L$ after release; beyond that it is frictionless. Air resistance is negligible. The cart's speed after traveling distance $L$ is $v$ . Which energy equation is correct?

$\tfrac12 kx^2 = \tfrac12 mv^2 + \mu_k mgL$ (correct answer)
$\tfrac12 kx^2 + \mu_k mgL = \tfrac12 mv^2$
$\tfrac12 kx^2 = \tfrac12 mv^2 - \mu_k mgL$
$\tfrac12 kx^2 = \tfrac12 m\vec v^2 + \mu_k mg\,\vec L$

Explanation: This question examines conservation of energy involving elastic potential, kinetic energy, and frictional work. The initial elastic potential energy is ½ k x², and friction does negative work -μ_k m g L over distance L. The energy balance is initial elastic potential plus work by friction equals final kinetic energy, leading to ½ k x² - μ_k m g L = ½ m v², or rearranged as ½ k x² = ½ m v² + μ_k m g L. This equation correctly captures the dissipation of energy due to friction. Choice C is incorrect because it subtracts the frictional term from the kinetic energy, reversing the energy loss. To solve similar problems, list all energy forms and subtract work by dissipative forces from the initial energy.

Question 5

A skier of mass $m$ descends a vertical drop $h$ and experiences air drag that dissipates energy $E_d$ . Starting from rest, what is the skier's speed at the bottom?

$v=\sqrt{2gh}$
$v=\sqrt{\dfrac{2(mgh-E_d)}{m}}$ (correct answer)
$v=\sqrt{\dfrac{2(mgh+E_d)}{m}}$
$v=\sqrt{2gh}-\sqrt{\dfrac{2E_d}{m}}$

Explanation: This problem involves energy conservation with energy dissipation by air drag. The energy equation is: initial gravitational potential energy equals final kinetic energy plus dissipated energy, or

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

. Rearranging to solve for

v

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

, so

v^2 = \frac{2(mgh - E_d)}{m}

, and therefore

v = \sqrt{\frac{2(mgh - E_d)}{m}}

. Choice C incorrectly adds

E_d

to the initial energy, which would mean air drag adds energy rather than removing it. When dealing with energy dissipation, always subtract the dissipated energy from the initial mechanical energy to find the remaining energy.

Question 6

A skier of mass $m$ descends vertical drop $h$ and then crosses a rough patch where friction does work $-W_f$ . What is the skier's kinetic energy after the patch?

$K = mgh - W_f$ (correct answer)
$K = mgh + W_f$
$K = m\vec g\cdot \vec h - W_f$
$K = mgh$ because total mechanical energy is conserved

Explanation: This problem requires energy accounting when both gravity and friction do work on a skier. The skier starts with gravitational potential energy

mgh

(taking the bottom as zero reference) and zero kinetic energy, then friction does negative work

-W_f

on the skier. The work-energy theorem states:

W_{\text{net}} = \Delta K

, where

W_{\text{net}} = W_{\text{gravity}} + W_{\text{friction}} = mgh + (-W_f) = mgh - W_f

. Since the skier starts from rest,

\Delta K = K_{\text{final}} - 0 = K_{\text{final}}

, so

K = mgh - W_f

. Choice B (

K = mgh + W_f

) is incorrect because it treats friction work as positive rather than negative—friction opposes motion and removes energy from the skier. When multiple forces do work, add their work algebraically, remembering that friction does negative work on moving objects.

Question 7

A block of mass $m$ is released from rest at height $h$ above the floor and slides down a rough ramp to the floor. The ramp exerts a constant kinetic friction force of magnitude $f_k$ over a distance $d$ along the ramp. Air resistance is negligible. The block reaches the bottom with speed $v$ . Which energy-accounting equation correctly relates these quantities?

$mgh - f_k d = \tfrac12 mv^2$ (correct answer)
$mgh + f_k d = \tfrac12 mv^2$
$mgh = \tfrac12 mv^2 + f_k\,\vec d$
$mg(h+d)=\tfrac12 mv^2$

Explanation: This question assesses the conservation of energy principle when non-conservative forces like friction are present. The initial gravitational potential energy of the block is mgh, which is partially converted to kinetic energy at the bottom and partially dissipated as thermal energy due to friction. The work done by friction is negative and equals -f_k d, so the energy equation is initial potential energy plus work by friction equals final kinetic energy. Thus, mgh - f_k d = ½ m v² correctly accounts for the energy transformation. Choice B is incorrect because it adds the frictional work instead of subtracting it, which would imply friction increases the kinetic energy. When applying conservation of energy with friction, always include the work done by non-conservative forces as part of the energy accounting to determine the final kinetic energy.

Question 8

A block of mass $m$ is released from rest at the top of a frictionless track at height $h$ above the bottom. At the bottom, it compresses a horizontal spring of constant $k$ by a maximum amount $x$ . Air resistance is negligible. Which equation correctly describes the energy transformation at maximum compression?

$mgh = \tfrac12 kx^2$ (correct answer)
$mgh + \tfrac12 kx^2 = 0$
$mgh = \tfrac12 kx^2\,\hat{x}$
$mg(h-x)=\tfrac12 kx^2$

Explanation: This question tests conservation of energy in a system converting gravitational to elastic potential. The block starts with gravitational potential mgh and ends at maximum compression with elastic potential ½ k x² and zero kinetic energy. Since the track is frictionless, mechanical energy is conserved, so initial potential equals final elastic potential. Therefore, mgh = ½ k x² is the correct relation. Choice D is incorrect as it uses (h - x), assuming x affects height, which it does not in a horizontal spring. For energy transformations, identify points of interest and equate energies while accounting only for conservative forces if no dissipation occurs.

Question 9

A cart of mass $m$ starts from rest at height $h$ above a reference level and rolls down a track. The track is smooth (friction negligible) until the cart enters a rough horizontal section of length $L$ , where kinetic friction of magnitude $f_k$ acts. The cart exits the rough section with speed $v$ . Which equation best represents conservation of energy with work by friction included?

$mgh = \tfrac12 mv^2 - f_k L$
$mgh = \tfrac12 mv^2 + f_k L$
$mgh - f_k L = \tfrac12 mv^2$ (correct answer)
$mgh = \tfrac12 m\vec v^2 + f_k L$

Explanation: This question evaluates understanding of conservation of energy with friction acting over a specific section of the path. The cart starts with gravitational potential energy mgh, which converts to kinetic energy, but friction does negative work -f_k L on the rough section. The conservation equation states that initial potential energy plus work by friction equals final kinetic energy after the rough section. Therefore, mgh - f_k L = ½ m v² properly relates the quantities. Choice B is incorrect as it adds f_k L to the kinetic energy, mistakenly treating frictional work as positive. A useful strategy is to identify conservative and non-conservative forces separately and ensure non-conservative work is subtracted when it opposes motion.

Question 10

A block of mass $m$ is released from rest at height $h$ above the floor on a frictionless track. At the bottom, it compresses a spring (spring constant $k$ ) on a horizontal surface. Friction and air resistance are negligible, and the spring is initially uncompressed. How far $x$ does the spring compress at maximum compression?

$x=\dfrac{mgh}{k}$
$x=\sqrt{\dfrac{2mgh}{k}}$ (correct answer)
$x=\sqrt{\dfrac{mgh}{2k}}$
$x=\dfrac{2mgh}{k}$

Explanation: This problem requires applying conservation of energy to find the spring compression distance. Initially, the block has gravitational potential energy mgh at height h and zero kinetic energy (released from rest). At maximum compression, the block momentarily stops (zero kinetic energy) and all energy is stored as elastic potential energy ½kx² in the spring. Setting initial energy equal to final energy: mgh = ½kx². Solving for x gives x = √(2mgh/k). Choice A incorrectly omits the factor of ½ from the spring's potential energy formula. The key strategy is to identify energy forms at initial and final states, then apply conservation when no non-conservative forces do work.

Question 11

A roller coaster car of mass $m$ moves along a track with negligible friction. Point $1$ is at height $h_1$ with speed $v_1$ , and point $2$ is at height $h_2$ with speed $v_2$ . Air resistance is negligible. Which relation between these quantities must be true?

$mgh_1 + \tfrac12 mv_1^2 = mgh_2 + \tfrac12 mv_2^2$ (correct answer)
$mgh_1 - \tfrac12 mv_1^2 = mgh_2 - \tfrac12 mv_2^2$
$m g (h_1-h_2)=\tfrac12 m(\vec v_2-\vec v_1)^2$
$mgh_1 + \tfrac12 mv_1^2 = 0$ if $U=0$ at point 2

Explanation: This question tests conservation of mechanical energy in a frictionless roller coaster system. Total mechanical energy at point 1 is m g h₁ + ½ m v₁². At point 2, it is m g h₂ + ½ m v₂². Since energy is conserved with no friction or air resistance, these totals are equal: m g h₁ + ½ m v₁² = m g h₂ + ½ m v₂². Choice B is incorrect as it subtracts kinetic terms, which would not preserve energy equality. Always sum potential and kinetic energies at different points and set them equal for conservative systems to find relationships between variables.

Question 12

A roller-coaster car of mass $m$ moves on a track with negligible friction. At point A it is at height $h_A$ with speed $v_A$ , and at point B it is at height $h_B$ with speed $v_B$ . Take $U_g=0$ at the ground. Which relation must be true?

$mgh_A+\tfrac12 mv_A^2=mgh_B+\tfrac12 mv_B^2$ (correct answer)
$mgh_A-\tfrac12 mv_A^2=mgh_B-\tfrac12 mv_B^2$
$\vec U_{g,A}+\vec K_A=\vec U_{g,B}+\vec K_B$
$mgh_A+mv_A=mgh_B+mv_B$

Explanation: This problem tests conservation of mechanical energy for a frictionless roller coaster. With no non-conservative forces (friction is negligible), total mechanical energy remains constant between any two points. At point A, the total energy is gravitational potential energy mgh_A plus kinetic energy ½mv_A². At point B, it's mgh_B + ½mv_B². Conservation of energy requires these totals to be equal: mgh_A + ½mv_A² = mgh_B + ½mv_B². Choice D incorrectly uses momentum (mv) instead of kinetic energy (½mv²) in the conservation equation. The strategy is to recognize when mechanical energy is conserved (no friction or air resistance) and write the conservation equation.

Question 13

A pendulum bob of mass $m$ is released from rest at angle such that its vertical drop to the lowest point is $\Delta h$ . Air resistance is negligible. At the lowest point, the bob collides with and sticks to a lump of clay of mass $m$ at rest, and the combined mass swings upward. During the collision, mechanical energy is not conserved, but momentum is. What is the maximum height $h_{\max}$ the stuck-together masses rise above the lowest point?

$h_{\max}=\Delta h$
$h_{\max}=\dfrac{\Delta h}{2}$
$h_{\max}=\dfrac{\Delta h}{4}$ (correct answer)
$h_{\max}=2\Delta h$

Explanation: This problem combines conservation of energy and conservation of momentum in a collision. First, the pendulum bob converts potential energy mgh into kinetic energy: mgh = ½mv₁², giving v₁ = √(2gh). During the perfectly inelastic collision, momentum is conserved: mv₁ = 2mv₂, so v₂ = v₁/2 = √(2gh)/2. The combined mass then rises, converting kinetic energy ½(2m)v₂² back to potential energy: ½(2m)v₂² = 2mgh_max. Substituting v₂: ½(2m)(gh/2) = 2mgh_max, which gives h_max = h/4. Choice A incorrectly assumes no energy is lost in the collision. The key is recognizing that momentum is conserved in the collision but mechanical energy is not.

Question 14

A block of mass $m$ slides down a vertical drop of height $h$ and then across a rough horizontal surface, coming to rest after distance $d$ . The only nonconservative force is kinetic friction on the horizontal surface; air resistance is negligible. Which expression gives the magnitude of the friction force $f_k$ ?

$f_k=\dfrac{\tfrac12 mv^2}{d}$ , where $v$ is the speed at the bottom
$f_k=\dfrac{mgh}{d}$ (correct answer)
$f_k=\dfrac{mgh}{2d}$
$\vec f_k=\dfrac{mgh}{d}\,\hat{x}$ because energy points along motion

Explanation: This problem requires finding the friction force using conservation of energy. The block starts with gravitational potential energy mgh at the top and zero kinetic energy. On the horizontal surface, friction does negative work W_f = -f_k·d, bringing the block to rest. The energy equation is: mgh + (-f_k·d) = 0, which gives f_k = mgh/d. Choice A incorrectly uses the kinetic energy at the bottom instead of recognizing that all initial potential energy is dissipated by friction. The key insight is that the work done by friction equals the total mechanical energy lost, which equals the initial potential energy.

Question 15

A ball rolls off a table of height $H$ with horizontal speed $v_0$ ; air resistance is negligible. Just before hitting the floor, what is its speed in terms of $v_0$ , $g$ , and $H$ ?

$v=\sqrt{v_0^2+2gH}$ (correct answer)
$v=v_0+\sqrt{2gH}$
$v=\sqrt{v_0^2-2gH}$
$v=\sqrt{2gH-v_0^2}$

Explanation: This question applies conservation of energy to a ball falling after rolling off a table. The initial kinetic energy is

(\frac{1}{2}) m v_0^2

(horizontal), and potential energy decreases by

m g H

. The vertical speed component is

\sqrt{2 g H}

, so total v =

\sqrt{v_0^2 + 2 g H}

. Energy is conserved separately in horizontal and vertical directions with no air resistance. Option B is incorrect as it adds speeds linearly instead of using vector magnitude. In combined motion problems, separate components and combine using Pythagoras for total speed before impact.

Question 16

A crate of mass $m$ is pulled up a ramp at constant speed by a rope over a distance $d$ along the ramp, gaining vertical height $h$ . Kinetic friction is present and does negative work of magnitude $f_k d$ . Air resistance is negligible. What is the work done by the rope, $W_T$ ?

$W_T = mgh - f_k d$
$W_T = mgh + f_k d$ (correct answer)
$W_T = mgh$ because constant speed implies no frictional losses
$\vec W_T = mgh\,\hat{y} + f_k d\,\hat{x}$

Explanation: This question examines work-energy principles for constant speed motion up a ramp with friction. Since speed is constant, change in kinetic energy is zero, and the net work equals zero. Work by gravity is

-mgh

, work by friction is

-f_k d

, so work by tension

W_T

must balance these:

W_T = mgh + f_k d

. This equation correctly accounts for overcoming both gravity and friction. Choice A is incorrect because it subtracts

f_k d

, implying tension does less work than needed. A transferable approach is to use the work-energy theorem, setting net work equal to

\Delta K

, and solve for unknown works.

Question 17

A ball of mass $m$ is thrown straight upward from height $y_0$ above the ground with initial speed $v_0$ . Air resistance is negligible. The ball reaches a maximum height $y_{\max}$ . Which equation correctly relates $y_{\max}$ to the given quantities if gravitational potential energy is zero at the ground?

$\tfrac12 mv_0^2 + mgy_0 = mgy_{\max}$ (correct answer)
$\tfrac12 mv_0^2 - mgy_0 = mgy_{\max}$
$\tfrac12 mv_0^2 = mg(y_{\max}-y_0)\,\hat{y}$
$\tfrac12 m\vec v_0^2 + mgy_0 = mgy_{\max}$

Explanation: This question assesses conservation of energy for projectile motion with potential energy zero at the ground. The ball starts with kinetic energy ½ m v₀² and potential energy m g y₀. At maximum height, kinetic energy is zero, so all mechanical energy is potential m g y_max. The conservation equation is ½ m v₀² + m g y₀ = m g y_max. Choice B is incorrect because it subtracts m g y₀, which would undervalue the initial energy contribution. A key strategy is to equate total mechanical energy at initial and final points, ensuring potential is calculated relative to the same reference.

Question 18

A pendulum bob of mass $m$ is released from rest at point $A$ and swings to the lowest point $B$ . Air resistance is negligible. The vertical drop from $A$ to $B$ is $\Delta y$ . A student sets the gravitational potential energy to zero at point $A$ instead of at $B$ . What is the correct expression for the bob's kinetic energy at $B$ ?

$K_B = mg\Delta y$ (correct answer)
$K_B = 0$ because $U_A=0$
$K_B = -mg\Delta y$
$K_B = mg\Delta y\,\hat{y}$

Explanation: This question tests the conservation of energy for a pendulum with an unconventional choice of zero potential energy. With gravitational potential set to zero at point A, the potential at B is -mg Δy since B is lower. Conservation of energy gives initial potential plus kinetic (both zero) equals final potential plus kinetic, so 0 = -mg Δy + K_B. Rearranging yields K_B = mg Δy as the kinetic energy at the bottom. Choice C is incorrect because it makes kinetic energy negative, which is impossible for a scalar quantity. Always verify the reference point for potential energy and adjust signs accordingly to maintain consistency in energy conservation problems.

Question 19

A ball of mass $m$ is thrown straight upward from height $y_0$ above the ground with initial speed $v_0$ . Air resistance is negligible. A student chooses gravitational potential energy to be zero at the launch point instead of at the ground. Which statement about the ball's maximum height above the ground is correct?

It depends on the chosen zero level for potential energy
It is smaller if $U_g=0$ is chosen at the launch point
It is larger if $U_g=0$ is chosen at the launch point
It is independent of the chosen zero level for potential energy (correct answer)

Explanation: This problem tests understanding of gravitational potential energy reference levels in conservation of energy. The ball's maximum height above ground depends only on its initial kinetic energy ½mv₀² converting to gravitational potential energy. Using energy conservation: ½mv₀² = mg(h_max - y₀), giving h_max = y₀ + v₀²/(2g). This result is independent of where we choose U_g = 0 because only differences in potential energy matter in conservation equations. Choice A incorrectly suggests the physical result depends on our arbitrary choice of reference level. The key insight is that energy differences, not absolute values, determine physical outcomes.

Question 20

Two identical balls are dropped from rest: Ball 1 from height $h$ and Ball 2 from height $2h$ . Air resistance is negligible. Just before hitting the ground, each ball has speed $v_1$ and $v_2$ , respectively. What is the ratio $v_2/v_1$ ?

$2$
$\sqrt{2}$ (correct answer)
$\dfrac{1}{\sqrt{2}}$
$\dfrac{1}{2}$

Explanation: This problem applies conservation of energy to free fall from different heights. For ball 1 dropped from height h: mgh = ½mv₁², giving v₁ = √(2gh). For ball 2 dropped from height 2h: mg(2h) = ½mv₂², giving v₂ = √(4gh) = 2√(gh). The ratio is v₂/v₁ = 2√(gh)/√(2gh) = 2/√2 = √2. Choice A incorrectly assumes speed is proportional to height rather than to the square root of height. The key insight is that kinetic energy is proportional to height, but speed is proportional to the square root of height.