AP Physics 1 Quiz: Potential Energy

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Potential EnergyQuestion 1 of 20

A block is released from rest while attached to a vertical spring. Define $U_s=0$ when the spring is unstretched. At instant $A$ the spring is stretched $0.08\,\text{m}$ ; at instant $B$ it is stretched $0.16\,\text{m}$ . Which statement about $U_s$ is correct?

U_s(A)>U_s(B)

because the spring force is smaller at

A

U_s(B)>U_s(A)

U_s(A)=U_s(B)

because both are stretches.

U_s(B)

is negative because the block moved downward.

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AP Physics 1 Quiz

AP Physics 1 Quiz: Potential Energy

Practice Potential 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 Potential 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

$U_s(A)>U_s(B)$ because the spring force is smaller at $A$ .
$U_s(B)>U_s(A)$ . (correct answer)
$U_s(A)=U_s(B)$ because both are stretches.
$U_s(B)$ is negative because the block moved downward.

Explanation: This question tests understanding of spring potential energy at different stretches. Spring potential energy is Us = ½kx², where x is the stretch from the unstretched position. At instant A, x = 0.08 m, so Us(A) = ½k(0.08)² = ½k(0.0064). At instant B, x = 0.16 m, so Us(B) = ½k(0.16)² = ½k(0.0256). Since 0.0256 = 4 × 0.0064, we have Us(B) = 4Us(A), so Us(B) > Us(A). The spring force being smaller at A doesn't mean higher potential energy—it means less stretch and lower energy. Spring potential energy is always positive for any stretch and doesn't depend on the direction of motion. When comparing spring energies, remember that doubling the displacement quadruples the potential energy.

Question 2

A mass on a horizontal spring is at position $P$ where the spring is compressed 0.20 m from equilibrium, and at position $Q$ where it is stretched 0.20 m. The reference is $U_s=0$ at equilibrium. Which is true?

$U_s(P)>U_s(Q)$ because compression stores more energy than stretch.
$U_s(P)=U_s(Q)$ . (correct answer)
$U_s(P)<U_s(Q)$ because the spring force is opposite.
$U_s(P)$ is negative while $U_s(Q)$ is positive.

Explanation: This question assesses elastic potential energy in AP Physics 1. Elastic potential energy depends on the magnitude of displacement from the equilibrium position, where U_s = 0. Whether compressed or stretched by the same amount, U_s = (1/2)k x^2 yields the same value since x is squared. The direction of displacement does not change the energy stored. Choice A is a distractor that wrongly assumes compression stores more energy than stretching, but both are equivalent in magnitude. Apply the squared displacement formula to ensure equal energies for symmetric positions.

Question 3

A block is held at rest against a vertical spring. The spring's natural length is defined as $U_s=0$ . At position 1 the spring is compressed 0.10 m; at position 2 it is compressed 0.30 m. Which is true?

$U_s(2)=3U_s(1)$ because compression tripled.
$U_s(2)>U_s(1)$ . (correct answer)
$U_s(2)<U_s(1)$ because the spring force is upward.
Spring potential energy is a vector, so direction matters.

Explanation: This question tests knowledge of elastic potential energy in AP Physics 1. The potential energy stored in a spring is based on its compression or extension from the natural length, defined as the reference where U_s = 0. For compressions, U_s = (1/2)k x^2, where x is the magnitude of displacement, so greater compression means higher energy. The direction of compression does not affect the scalar value of energy. Choice A incorrectly assumes a linear relationship, but energy scales with the square of compression, making it nine times greater, not three. Always use the formula U_s = (1/2)k x^2 to compare energies in different spring configurations.

Question 4

A horizontal spring (spring constant $k$ ) is fixed to a wall. The spring's unstretched length is marked as position $x=0$ . A cart is held at rest at position $x=+0.10\,\text{m}$ (spring stretched) and then moved slowly to $x=-0.10\,\text{m}$ (spring compressed). The reference for spring potential energy is $U_s=0$ at $x=0$ .

At which position is the spring potential energy greater?

At $x=+0.10\,\text{m}$ only, because stretching stores energy but compression does not.
At $x=-0.10\,\text{m}$ only, because the spring force is opposite the displacement.
They are equal at both positions, since $U_s$ depends on $x^2$ . (correct answer)
Cannot be determined without knowing the cart's mass.

Explanation: This question assesses understanding of elastic potential energy in AP Physics 1. Elastic potential energy is determined by the spring's displacement from its equilibrium position, where U_s = 0 is set at x = 0. The formula U_s = (1/2)kx^2 shows that energy depends on the square of displacement, making it the same for equal magnitudes regardless of direction. Thus, at x = +0.10 m and x = -0.10 m, the potential energies are equal since both have |x| = 0.10 m. A common distractor, choice A, wrongly claims energy is stored only in stretching, not compression, but the formula applies to both. To compare elastic potential energies, focus on the magnitude of displacement from the reference equilibrium position.

Question 5

A spring on a horizontal table has spring constant $k$ . The spring's unstretched length is the reference where $U_s=0$ . At state $A$ the spring is stretched by $x$ , and at state $B$ it is stretched by $2x$ . Which comparison is correct?

$U_s(B)=4U_s(A)$ . (correct answer)
$U_s(B)=2U_s(A)$ because the stretch doubled.
$U_s(B)=U_s(A)$ because both are stretched.
$U_s(B)$ is negative because the spring pulls inward.

Explanation: This question evaluates understanding of elastic potential energy in AP Physics 1. Elastic potential energy is determined by the displacement of the spring from its equilibrium position, where U_s = 0 is often set at the unstretched length. The energy is given by U_s = (1/2)kx^2, so it increases quadratically with the displacement x, regardless of direction. Doubling the stretch quadruples the energy because of the squared term. Choice B is a distractor that mistakenly assumes a linear relationship instead of quadratic. A useful strategy is to recall that elastic potential energy depends on the square of the displacement for transferable applications to springs.

Question 6

A book is lifted slowly in Earth's gravitational field. The reference level is chosen at the tabletop, so $U_g=0$ at the tabletop. The book starts on the floor at $h=-0.80\,\text{m}$ relative to the tabletop and ends on a shelf at $h=+0.40\,\text{m}$ . Which statement about $U_g$ is correct?

The book's gravitational potential energy is always positive during the lift.
$U_g$ on the floor is greater than $U_g$ on the shelf because the floor is lower.
$U_g$ on the floor is negative relative to this reference level. (correct answer)
$U_g$ is a vector, so it points upward on the shelf.

Explanation: This question tests understanding of potential energy with different reference levels. When the reference level is at the tabletop (Ug = 0 there), positions below the tabletop have negative heights and therefore negative potential energies. The book starts at h = -0.80 m (below the reference), so Ug = mg(-0.80) is negative. The book ends at h = +0.40 m (above the reference), so Ug = mg(0.40) is positive. Option B incorrectly suggests that lower positions have greater potential energy—they actually have lower (more negative) values. Potential energy is a scalar, not a vector, so it doesn't point in any direction. Remember that potential energy can be negative when the object is below the chosen reference level.

Question 7

A cart is on a frictionless track in Earth's gravitational field. The reference level is the floor ( $U_g=0$ at $h=0$ ). At point $P$ , the cart is at height $h=2.0\,\text{m}$ . At point $Q$ , the cart is at height $h=5.0\,\text{m}$ . Which statement about gravitational potential energy is correct?

$U_g(P)>U_g(Q)$ because the cart is closer to the reference level at $P$ .
$U_g(P)=U_g(Q)$ because the force of gravity is constant.
$U_g(Q)>U_g(P)$ because $Q$ is at greater height above the reference level. (correct answer)
$\vec{U}_g(Q)$ points upward and is larger than $\vec{U}_g(P)$ .

Explanation: This question tests understanding of gravitational potential energy. Gravitational potential energy is given by Ug = mgh, where h is the height above a chosen reference level. Since point Q (h = 5.0 m) is higher than point P (h = 2.0 m) above the reference level (floor), the potential energy at Q is greater than at P. The fact that gravity is constant doesn't make the potential energies equal—it means the formula Ug = mgh applies consistently. Potential energy is a scalar quantity, not a vector, so option D is incorrect. When comparing potential energies, always identify which position is higher above the reference level.

Question 8

A ball is at rest at two locations in a uniform gravitational field. The reference level is set at location $X$ , so $U_g(X)=0$ . Location $Y$ is $1.0\,\text{m}$ below $X$ . Which statement about $U_g(Y)$ is correct?

$U_g(Y)$ is negative relative to this reference level. (correct answer)
$U_g(Y)=0$ because gravity is conservative.
$U_g(Y)$ is positive because potential energy is always positive.
$\vec{U}_g(Y)$ points downward because $Y$ is below $X$ .

Explanation: This question tests understanding of gravitational potential energy below a reference level. With location X as the reference (Ug(X) = 0), any location below X has negative potential energy. Location Y is 1.0 m below X, which means h = -1.0 m relative to X, so Ug(Y) = mg(-1.0) = -mg, which is negative. Gravity being conservative doesn't make the potential energy zero—it means the energy depends only on position, not path. Potential energy can be negative when below the reference level; it's not always positive. Potential energy is a scalar quantity, not a vector. When an object is below your chosen reference level, its gravitational potential energy is negative.

Question 9

A block is attached to a spring on a table. The spring constant is larger in setup 1 than setup 2 ( $k_1>k_2$ ). In both setups, the spring is stretched the same distance $x$ from equilibrium ( $U_s=0$ at equilibrium). Which is true?

$U_{s,1}>U_{s,2}$ . (correct answer)
$U_{s,1}=U_{s,2}$ because the stretch is the same.
$U_{s,1}<U_{s,2}$ because the larger $k$ means less displacement.
Potential energy points along the spring force, so larger $k$ reverses its direction.

Explanation: This question tests understanding of elastic potential energy in AP Physics 1. The energy stored relates to the spring constant k and displacement x from equilibrium, where U_s = 0. A larger k means more energy for the same x because U_s = (1/2)k x^2 scales with k. Displacement is fixed, so energy differs based on k. Choice B incorrectly assumes equal energies for equal stretches, ignoring k's role. Compare setups by factoring in both k and x for potential energy differences.

Question 10

A block slides on a frictionless track in a uniform gravitational field. Point $A$ is at height $h=1.0\,\text{m}$ above the ground, and point $B$ is at height $h=3.0\,\text{m}$ . A student chooses the reference level so that $U_g=0$ at point $B$ .

With this reference choice, which statement is correct?

$U_g(A)$ is negative. (correct answer)
$U_g(A)$ must be positive because potential energy cannot be negative.
$U_g(A)=0$ because $A$ is on the track.
$U_g(B)$ is the largest because $B$ is the highest point.

Explanation: This question assesses understanding of gravitational potential energy in AP Physics 1. Gravitational potential energy is defined relative to a chosen reference point where U_g = 0, which can lead to negative values below that point. Here, the reference is at point B (3.0 m), so U_g(B) = 0, and point A is 2.0 m below it at 1.0 m. Thus, U_g(A) = mg(-2.0 m) is negative, reflecting a lower position relative to the reference. A common distractor, choice B, insists potential energy cannot be negative, but it can depending on reference choice. To handle unusual references, calculate signed heights from the zero point for accurate potential energy values.

Question 11

A mass $m$ is attached to a vertical spring and held at rest. The spring's unstretched length position is labeled $x=0$ . Position $A$ corresponds to the spring stretched downward by $0.05\,\text{m}$ , and position $B$ corresponds to the spring stretched downward by $0.10\,\text{m}$ . The reference for spring potential energy is $U_s=0$ at $x=0$ .

Which comparison of spring potential energies is correct?

$U_s(A)>U_s(B)$ because the spring force is smaller at $A$ .
$U_s(A)=U_s(B)$ because both positions are below $x=0$ .
$U_s(B)>U_s(A)$ because the stretch magnitude is larger at $B$ . (correct answer)
Cannot be determined without knowing $m$ .

Explanation: This question assesses understanding of elastic potential energy in AP Physics 1. Elastic potential energy is determined by the displacement from the unstretched position where U_s = 0. The energy U_s = (1/2)kx^2 increases with larger displacement magnitudes, regardless of direction. Here, position B has a larger stretch (0.10 m) than A (0.05 m), so U_s(B) > U_s(A). A common distractor, choice A, reverses this by linking energy to force magnitude incorrectly, but energy scales with x^2. To compare spring potential energies, compute (1/2)kx^2 using displacements from the reference equilibrium.

Question 12

A block is held at rest in a uniform gravitational field. Point $A$ is on a shelf $2.0\,\text{m}$ above the floor, and point $B$ is on the floor. The reference level for gravitational potential energy is chosen as $U_g=0$ at the floor. The block is moved slowly from $B$ to $A$ .

Which statement about the gravitational potential energy is correct?

$U_g(A)<U_g(B)$ because gravity points downward.
$U_g(A)>U_g(B)$ because $A$ is higher than the reference level. (correct answer)
$U_g(A)=U_g(B)$ because potential energy is a vector and only direction changes.
$U_g(A)=0$ because potential energy must be zero at the highest point.

Explanation: This question assesses understanding of gravitational potential energy in AP Physics 1. Gravitational potential energy is determined by an object's position relative to a chosen reference level where U_g = 0. In this case, the reference is at the floor, so point B has zero potential energy, while point A is 2.0 m above it. Thus, U_g at A is positive and greater than at B, as potential energy increases with height above the reference. A common distractor, choice A, incorrectly suggests U_g(A) < U_g(B) due to gravity's direction, but potential energy is scalar and depends on height, not force direction. To compare gravitational potential energies, always calculate heights relative to the defined reference point.

Question 13

A pendulum bob is in a uniform gravitational field. Point $A$ is the lowest point of the swing, and point $B$ is $0.30\,\text{m}$ vertically above $A$ along the bob's path. The reference level is chosen so that $U_g=0$ at point $A$ . The bob is moved slowly from $A$ to $B$ .

Which comparison of gravitational potential energy is correct?

$U_g(B)<U_g(A)$ because the bob is farther from Earth at $B$ .
$U_g(B)>U_g(A)$ because $B$ is above the reference level. (correct answer)
$U_g(B)=U_g(A)$ because only kinetic energy changes in a pendulum.
$U_g(B)$ must be negative because gravity is an attractive force.

Explanation: This question assesses understanding of gravitational potential energy in AP Physics 1. Gravitational potential energy depends on the vertical position relative to the chosen reference where U_g = 0. Here, the reference is at point A, the lowest point, so U_g(A) = 0, and point B is 0.30 m above it. Therefore, U_g(B) is positive and greater than U_g(A), as energy increases with height above the reference. A common distractor, choice A, reverses the inequality by miscounting the height difference, but B is clearly above A. To compare gravitational potential energies in pendulums, measure vertical heights from the defined reference level.

Question 14

A block is attached to a horizontal spring on a frictionless table. The spring's unstretched length is defined as $x=0$ where $U_s=0$ . At position $A$ , the spring is compressed $0.10\,\text{m}$ . At position $B$ , it is compressed $0.20\,\text{m}$ . Which comparison of spring potential energies is correct?

$U_s(B)>U_s(A)$ . (correct answer)
$U_s(A)>U_s(B)$ because the spring force is smaller at $A$ .
$U_s(A)=U_s(B)$ because both positions are compressions.
$U_s(B)$ is negative because the spring is compressed.

Explanation: This question tests understanding of spring potential energy. Spring potential energy is given by Us = ½kx², where x is the displacement from equilibrium (unstretched position). At position A, the spring is compressed 0.10 m, giving Us(A) = ½k(0.10)². At position B, the spring is compressed 0.20 m, giving Us(B) = ½k(0.20)² = ½k(0.04) = 4 × ½k(0.01) = 4Us(A). Since (0.20)² = 0.04 is greater than (0.10)² = 0.01, the potential energy at B is greater than at A. The direction of compression doesn't matter—only the magnitude of displacement from equilibrium determines the potential energy. Spring potential energy is always positive when the spring is displaced from equilibrium, regardless of whether it's compressed or stretched.

Question 15

A 2.0 kg mass hangs from a vertical spring at rest. Define $U_s=0$ when the spring is unstretched. At position $R$ , the spring is stretched $0.05\,\text{m}$ . At position $S$ , it is stretched $0.15\,\text{m}$ . Which statement about spring potential energy is correct?

$U_s(R)=U_s(S)$ because both are stretches, not compressions.
$U_s(S)>U_s(R)$ . (correct answer)
$U_s(S)$ is smaller because the spring force is downward at $S$ .
$U_s(R)$ must be negative because gravity is present.

Explanation: This question tests understanding of spring potential energy for different stretches. Spring potential energy is Us = ½kx², where x is the displacement from the unstretched position. At position R, x = 0.05 m, so Us(R) = ½k(0.05)². At position S, x = 0.15 m, so Us(S) = ½k(0.15)². Since (0.15)² = 0.0225 is greater than (0.05)² = 0.0025, we have Us(S) > Us(R). The fact that both positions involve stretching (not compression) doesn't make the energies equal—the amount of stretch matters. Spring potential energy depends only on the magnitude of displacement, not on the direction of the spring force or the presence of gravity. When comparing spring potential energies, square the displacements and compare those values.

Question 16

Two identical masses are held at rest in Earth's gravitational field. The reference level is the ground ( $U_g=0$ at ground). Mass 1 is at height $1.5\,\text{m}$ ; Mass 2 is at height $0.5\,\text{m}$ . Which comparison is correct?

$U_{g,1}=U_{g,2}$ because both experience the same $g$ .
$U_{g,2}>U_{g,1}$ because it is closer to the ground.
$U_{g,1}>U_{g,2}$ . (correct answer)
Both potential energies must be negative since gravity pulls downward.

Explanation: This question tests understanding of gravitational potential energy for objects at different heights. With the ground as reference (Ug = 0), gravitational potential energy is Ug = mgh. Mass 1 at height 1.5 m has Ug,1 = mg(1.5), while Mass 2 at height 0.5 m has Ug,2 = mg(0.5). Since 1.5 > 0.5 and the masses are identical, Ug,1 > Ug,2. The fact that both experience the same g doesn't make their potential energies equal—their different heights matter. Being closer to the ground means lower (not higher) potential energy. With the ground as reference, both potential energies are positive, not negative. To compare gravitational potential energies, identify which object is higher above the reference level.

Question 17

A block attached to a spring is held on a frictionless surface. Define $U_s=0$ at $x=0$ . At position 1, the spring is stretched $0.30\,\text{m}$ . At position 2, the spring is stretched $0.10\,\text{m}$ . Which statement is correct?

$U_s(2)>U_s(1)$ because the spring force is smaller at position 2.
$U_s(1)>U_s(2)$ . (correct answer)
$U_s(1)=U_s(2)$ because both displacements are positive.
$U_s(1)$ is negative because the spring pulls back toward equilibrium.

Explanation: This question tests understanding of spring potential energy at different stretches. Spring potential energy is Us = ½kx², where x is the displacement from equilibrium. At position 1, x = 0.30 m, so Us(1) = ½k(0.30)² = ½k(0.09). At position 2, x = 0.10 m, so Us(2) = ½k(0.10)² = ½k(0.01). Since 0.09 > 0.01, we have Us(1) > Us(2). The spring force being smaller at position 2 (less stretch) corresponds to lower potential energy, not higher. Both displacements being positive doesn't make the energies equal—the magnitude matters. Spring potential energy is always positive for any non-zero displacement. To compare spring potential energies, square the displacements and see which is larger.

Question 18

A block is attached to a horizontal spring. At point $A$ the spring is compressed by $x=0.20\,\text{m}$ , and at point $B$ the spring is compressed by $x=0.10\,\text{m}$ . A student chooses a reference such that $U_s=0$ when the spring is compressed by $0.10\,\text{m}$ .

With this reference, which statement is correct?

$U_s(A)>0$ . (correct answer)
$U_s(A)=0$ because $A$ is more compressed.
$U_s(A)<0$ because compression makes spring potential energy negative.
$U_s(B)>U_s(A)$ because the spring force is smaller at $B$ .

Explanation: This question assesses understanding of elastic potential energy in AP Physics 1. Elastic potential energy can be referenced to any point, here set to U_s = 0 at 0.10 m compression, allowing positive or negative values elsewhere. At point A (0.20 m compression), the larger displacement gives higher (1/2)kx^2 than at 0.10 m, so relative U_s(A) > 0. Point B is at the reference, so U_s(B) = 0. A common distractor, choice C, claims compression yields negative energy, but signs depend on the chosen reference. To work with shifted references, subtract the reference energy from the standard (1/2)kx^2 formula.

Question 19

A cart is on a frictionless incline in a uniform gravitational field. Point $A$ and point $B$ are at the same vertical height above the ground, but $A$ is on a steeper section where the incline angle is larger. The reference level is chosen so that $U_g=0$ at the ground.

Which statement about gravitational potential energy is correct?

$U_g(A)>U_g(B)$ because the component of weight along the incline is larger at $A$ .
$U_g(A)=U_g(B)$ because they are at the same height. (correct answer)
$U_g(A)<U_g(B)$ because the incline is steeper at $A$ .
$U_g$ is a vector, so $U_g(A)$ and $U_g(B)$ differ in direction.

Explanation: This question assesses understanding of gravitational potential energy in AP Physics 1. Gravitational potential energy depends solely on vertical height above the reference level, not on path or incline angle. Here, points A and B are at the same height above the ground where U_g = 0, so their potential energies are equal. The steeper incline at A does not affect this, as only vertical position matters. A common distractor, choice A, incorrectly ties energy to the weight component along the incline, but potential is path-independent. To compare potential energies on inclines, measure vertical heights from the reference, ignoring slope variations.

Question 20

A cart moves along a frictionless hill in a uniform gravitational field. The reference level is set at point $M$ so that $U_g(M)=0$ . Point $N$ is $3.0\,\text{m}$ above $M$ . Which statement about gravitational potential energy at $N$ is correct?

$U_g(N)=0$ because potential energy depends only on the path taken.
$U_g(N)$ is negative because the cart is above the reference point.
$U_g(N)$ is greater than $U_g(M)$ . (correct answer)
$\vec{U}_g(N)$ points upward and has magnitude $mg(3.0\,\text{m})$ .

Explanation: This question tests understanding of gravitational potential energy relative to a reference point. When point M is chosen as the reference level (Ug = 0), any point above M has positive potential energy. Point N is 3.0 m above M, so Ug(N) = mg(3.0 m), which is positive and greater than Ug(M) = 0. Potential energy doesn't depend on the path taken—only on the vertical position relative to the reference. Potential energy is a scalar quantity, not a vector, so it doesn't point in any direction. When a reference level is chosen, positions above it have positive potential energy and positions below it have negative potential energy.