Potential Energy
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AP Physics 1 › Potential Energy
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 $A$.
$U_s(B)$ is negative because the block moved downward.
$U_s(B)>U_s(A)$.
$U_s(A)=U_s(B)$ because both are stretches.
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.
A spring is compressed $0.10,\text{m}$ from its relaxed length. Taking $U_s=0$ at $x=0$, which statement is correct?
$U_s$ points opposite the compression direction.
$U_s$ is larger if the spring force is larger at that instant.
$U_s<0$ because compression is negative $x$.
$U_s>0$.
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.
Two points on a cliff are labeled $A$ and $B$. The reference level is chosen so $U_g=0$ at point $A$. Point $B$ is 4.0 m below $A$. Which is correct?
$U_g(B)=0$ because potential energy cannot be negative.
$U_g(B)$ points downward, so it is negative.
$U_g(B)>0$ because the ball is closer to Earth.
$U_g(B)<0$.
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.
A book is on a shelf. The reference level is chosen at the tabletop so $U_g=0$ there. The shelf is 0.80 m above the tabletop. Which statement is correct?
The book’s $U_g$ is a vector pointing upward.
The book’s $U_g$ is positive relative to the tabletop reference.
The book’s $U_g$ is negative because it could fall.
The book’s $U_g$ is zero because it is not moving.
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.
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 the spring force is opposite.
$U_s(P)=U_s(Q)$.
$U_s(P)>U_s(Q)$ because compression stores more energy than stretch.
$U_s(P)$ 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.
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?
$U_s(2)<U_s(1)$ because the spring force is upward.
$U_s(2)=3U_s(1)$ because compression tripled.
$U_s(2)>U_s(1)$.
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.
A 0.50 kg cart rests on a frictionless track. Taking $U_g=0$ at point $R$, the cart is at point $P$ that is 2.0 m above $R$. Which statement about gravitational potential energy is correct?
$U_g$ is a vector pointing upward at $P$.
$U_g(P)>U_g(R)$.
$U_g(P)=0$ because the cart is at rest.
$U_g(P)<U_g(R)$ because gravity points downward.
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.
A spring is compressed $x_1=0.04,\text{m}$ and then compressed to $x_2=0.08,\text{m}$. With $U_s=0$ at $x=0$, which is true?
$U_s$ is a vector so it cannot be compared without directions.
$U_s(x_2)>U_s(x_1)$.
$U_s(x_2)=2U_s(x_1)$ because the compression doubled.
$U_s(x_2)<U_s(x_1)$ because the spring force points opposite $x$.
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.
A cart is at point $P$, $1.5,\text{m}$ above point $Q$. Define $U_g=0$ at $Q$. Which is true?
$U_g(P)>U_g(Q)$.
$U_g$ at each point depends on the cart’s speed.
$U_g(P)<U_g(Q)$ because gravity is downward.
$U_g(P)=U_g(Q)$ because only height changes.
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.
A book rests on a shelf $2.0,\text{m}$ above the floor, where $U_g=0$. What is true about its gravitational potential energy?
$U_g=0$ because the book is at rest.
$U_g>0$.
$U_g<0$ because gravity acts downward.
$U_g$ is a vector pointing upward.
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.