AP Physics 1 Quiz: Potential Energy

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.

Question 1

1. $U_s(A)>U_s(B)$ because the spring force is smaller at $A$.
2. $U_s(B)>U_s(A)$. (correct answer)
3. $U_s(A)=U_s(B)$ because both are stretches.
4. $U_s(B)$ 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

1. $U_s<0$ because compression is negative $x$.
2. $U_s$ is larger if the spring force is larger at that instant.
3. $U_s>0$. (correct answer)
4. $U_s$ points opposite the compression direction.

Explanation: This question tests understanding of elastic potential energy. Elastic potential energy is given by $U_s = \frac{1}{2}kx^2$, where $x$ is the displacement from the relaxed position. Since the spring is compressed by $0.10\,\text{m}$, we have $x = -0.10\,\text{m}$ (negative for compression), but $x^2 = 0.01\,\text{m}^2$ is always positive. Therefore, $U_s = \frac{1}{2}k(0.01) > 0$ regardless of whether the spring is compressed or stretched. Choice A incorrectly assumes that compression makes the energy negative, but the squared term ensures positive energy. The strategy is to remember that elastic potential energy depends on the square of displacement, making it always positive for any non-zero displacement.

Question 3

1. $U_g(B)=0$ because potential energy cannot be negative.
2. $U_g(B)>0$ because the ball is closer to Earth.
3. $U_g(B)<0$. (correct answer)
4. $U_g(B)$ points downward, so it is negative.

Explanation: This question explores gravitational potential energy in AP Physics 1. The value of U_g at a point is set by its vertical position compared to the arbitrary reference where U_g = 0. Positions below the reference have negative h, resulting in negative U_g = mgh. This negativity indicates the system has less potential energy than at the reference. Choice A incorrectly claims potential energy cannot be negative, but it can depending on the reference choice. Remember to allow for negative values when the position is below the reference for accurate energy comparisons.

Question 4

1. The book’s $U_g$ is negative because it could fall.
2. The book’s $U_g$ is zero because it is not moving.
3. The book’s $U_g$ is positive relative to the tabletop reference. (correct answer)
4. The book’s $U_g$ is a vector pointing upward.

Explanation: This question examines gravitational potential energy in AP Physics 1. U_g is defined relative to a reference like the tabletop where it is zero. Positions above this reference have positive U_g = mgh, with h being the height difference. The value is positive and depends only on this relative position, not motion. Choice A is a distractor that confuses potential for falling with actual negative energy, but it's positive above the reference. Select a reference and compute height differences for consistent U_g evaluations.

Question 5

1. $U_s(P)>U_s(Q)$ because compression stores more energy than stretch.
2. $U_s(P)=U_s(Q)$. (correct answer)
3. $U_s(P)<U_s(Q)$ because the spring force is opposite.
4. $U_s(P)$ 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 6

1. $U_s(2)=3U_s(1)$ because compression tripled.
2. $U_s(2)>U_s(1)$. (correct answer)
3. $U_s(2)<U_s(1)$ because the spring force is upward.
4. 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 7

1. $U_g(P)=0$ because the cart is at rest.
2. $U_g(P)>U_g(R)$. (correct answer)
3. $U_g(P)<U_g(R)$ because gravity points downward.
4. $U_g$ is a vector pointing upward at $P$.

Explanation: This question assesses the concept of gravitational potential energy in AP Physics 1. Gravitational potential energy depends on an object's vertical position relative to a chosen reference point where it is defined as zero. At a higher position above the reference, the potential energy is positive and greater because height h in U_g = mgh is larger. At the reference point, U_g is zero, so any point above has higher potential energy. Choice A is incorrect because potential energy does not depend on whether the object is at rest or moving. To compare potential energies, always identify the reference point and measure heights relative to it.

Question 8

1. $U_s(x_2)=2U_s(x_1)$ because the compression doubled.
2. $U_s(x_2)>U_s(x_1)$. (correct answer)
3. $U_s(x_2)<U_s(x_1)$ because the spring force points opposite $x$.
4. $U_s$ is a vector so it cannot be compared without directions.

Explanation: This question tests understanding of how elastic potential energy scales with displacement. Elastic potential energy is $U_s = \frac{1}{2}kx^2$, so it varies with the square of displacement. At $x_1 = 0.04\,\text{m}$, $U_s(x_1) = \frac{1}{2}k(0.04)^2 = 0.0008k$. At $x_2 = 0.08\,\text{m}$, $U_s(x_2) = \frac{1}{2}k(0.08)^2 = 0.0032k = 4 \times U_s(x_1)$. Therefore, $U_s(x_2) > U_s(x_1)$, and specifically, the energy quadruples when displacement doubles. Choice A incorrectly assumes linear scaling, but potential energy scales quadratically with displacement. The key insight is that elastic potential energy increases with the square of displacement, not linearly.

Question 9

1. $U_g(P)>U_g(Q)$. (correct answer)
2. $U_g(P)=U_g(Q)$ because only height changes.
3. $U_g(P)<U_g(Q)$ because gravity is downward.
4. $U_g$ at each point depends on the cart’s speed.

Explanation: This question tests understanding of gravitational potential energy at different heights. Gravitational potential energy is $U_g = mgh$, where $h$ is measured from the reference level. Point P is $1.5\,\text{m}$ above point Q, and since $U_g = 0$ at Q, we have $U_g(P) = mg(1.5) > 0$ while $U_g(Q) = 0$. Therefore, $U_g(P) > U_g(Q)$. Choice C incorrectly thinks that the downward direction of gravity makes higher positions have lower potential energy, but higher positions always have greater gravitational potential energy. The key insight is that gravitational potential energy increases with height above the reference level, regardless of the direction of the gravitational force.

Question 10

1. $U_g$ is a vector pointing upward.
2. $U_g>0$. (correct answer)
3. $U_g<0$ because gravity acts downward.
4. $U_g=0$ because the book is at rest.

Explanation: This question tests understanding of gravitational potential energy. Gravitational potential energy is defined as $U_g = mgh$, where $h$ is the height above a chosen reference level. Since the book is at height $h = 2.0\,\text{m}$ above the floor (where $U_g = 0$), we have $U_g = mg(2.0) > 0$. Choice D incorrectly assumes that being at rest means zero potential energy, but potential energy depends only on position, not motion. The key strategy is to remember that potential energy is a scalar quantity determined by position relative to the reference level, and objects above the reference have positive potential energy.

Question 11

1. At $x=+0.10\,\text{m}$ only, because stretching stores energy but compression does not.
2. At $x=-0.10\,\text{m}$ only, because the spring force is opposite the displacement.
3. They are equal at both positions, since $U_s$ depends on $x^2$. (correct answer)
4. 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 12

1. $U_s(B)=4U_s(A)$. (correct answer)
2. $U_s(B)=2U_s(A)$ because the stretch doubled.
3. $U_s(B)=U_s(A)$ because both are stretched.
4. $U_s(B)$ 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 13

1. smaller because $x$ is negative
2. greater (correct answer)
3. zero because compression stores no energy
4. a leftward vector proportional to $x^2$

Explanation: This question tests understanding of spring potential energy under compression. Spring potential energy is Us = ½kx², where x is displacement from natural length. At x = -0.12 m (compression), Us = ½k(-0.12)² = ½k(0.0144) > 0, while at x = 0, Us = 0. The negative sign of x disappears when squared, so compression gives the same positive energy as equal extension. Since any positive value is greater than zero, the compressed spring has greater potential energy. Choice A incorrectly assumes negative x gives smaller energy, missing that x² is always positive. The key principle is that spring potential energy depends on displacement squared, making it positive for any non-zero displacement.

Question 14

1. $U_g<0$
2. $U_g$ is a vector pointing upward
3. $U_g>0$ (correct answer)
4. $U_g=0$ because the book is at rest

Explanation: This question tests understanding of gravitational potential energy relative to a reference point. Gravitational potential energy is calculated as Ug = mgh, where h is the height above the reference level where Ug = 0. Since the book is 2.0 m above the floor (our reference where Ug = 0), and height is positive, the potential energy must be positive: Ug = mg(2.0) > 0. Choice D incorrectly assumes that being at rest means zero potential energy, but potential energy depends on position, not motion. When an object is above your chosen zero reference level, its gravitational potential energy is always positive.

Question 15

1. negative because compression is in the negative direction
2. greater (correct answer)
3. zero because the spring force is internal
4. a leftward vector proportional to $x$

Explanation: This question tests understanding of elastic potential energy in springs. Spring potential energy is calculated as Us = ½kx², where x is the displacement from natural length (where Us = 0). Whether the spring is compressed or stretched, x² is always positive, making Us always positive for any non-zero displacement. At x = 0.10 m compression, Us = ½k(0.10)² > 0, while at natural length (x = 0), Us = 0. Choice A incorrectly assumes compression gives negative energy, but the squared term ensures positive energy. The key insight is that spring potential energy depends on x² not x, so it's always positive regardless of compression or extension direction.

Question 16

1. The book’s gravitational potential energy is always positive during the lift.
2. $U_g$ on the floor is greater than $U_g$ on the shelf because the floor is lower.
3. $U_g$ on the floor is negative relative to this reference level. (correct answer)
4. $U_g$ 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 17

1. $U_g(P)>U_g(Q)$ because the cart is closer to the reference level at $P$.
2. $U_g(P)=U_g(Q)$ because the force of gravity is constant.
3. $U_g(Q)>U_g(P)$ because $Q$ is at greater height above the reference level. (correct answer)
4. $\vec{U}_g(Q)$ 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 18

1. $U_g(Y)$ is negative relative to this reference level. (correct answer)
2. $U_g(Y)=0$ because gravity is conservative.
3. $U_g(Y)$ is positive because potential energy is always positive.
4. $\vec{U}_g(Y)$ 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 19

1. $U_{s,1}>U_{s,2}$. (correct answer)
2. $U_{s,1}=U_{s,2}$ because the stretch is the same.
3. $U_{s,1}<U_{s,2}$ because the larger $k$ means less displacement.
4. 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 20

1. $U_g$ is positive at the ball. (correct answer)
2. $U_g$ is negative at the ball because gravity is downward.
3. $U_g$ is zero because only changes in $U_g$ matter.
4. $U_g$ is a vector pointing upward.

Explanation: This question examines gravitational potential energy concepts in AP Physics 1. Gravitational potential energy is determined by an object's height relative to a reference level where U_g = 0. If the object is above the reference, h is positive, leading to positive U_g = mgh. The sign and value depend solely on this relative position, not on the direction of gravity. Choice B is a common distractor that confuses the direction of the gravitational force with the scalar nature of potential energy. To solve similar problems, choose a consistent reference point and calculate heights accordingly.