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?
$U_s(B)$ is negative because the block moved downward.
$U_s(A)=U_s(B)$ because both are stretches.
$U_s(A)>U_s(B)$ because the spring force is smaller at $A$.
$U_s(B)>U_s(A)$.
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 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 force of gravity is constant.
$U_g(P)>U_g(Q)$ because the cart is closer to the reference level at $P$.
$U_g(Q)>U_g(P)$ because $Q$ is at greater height above the reference level.
$\vec{U}_g(Q)$ 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.
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.
$\vec{U}_g(Y)$ points downward because $Y$ is below $X$.
$U_g(Y)$ is positive because potential energy is always positive.
$U_g(Y)=0$ because gravity is conservative.
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.
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?
$U_g$ on the floor is negative relative to this reference level.
The book’s gravitational potential energy is always positive during the lift.
$U_g$ is a vector, so it points upward on the shelf.
$U_g$ on the floor is greater than $U_g$ on the shelf because the floor is lower.
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.
A spring is compressed $0.10,\text{m}$ from its relaxed length. Taking $U_s=0$ at $x=0$, which statement is correct?
$U_s<0$ because compression is negative $x$.
$U_s$ is larger if the spring force is larger at that instant.
$U_s$ points opposite the compression direction.
$U_s>0$.
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 zero because it is not moving.
The book’s $U_g$ is a vector pointing upward.
The book’s $U_g$ is negative because it could fall.
The book’s $U_g$ is positive relative to the tabletop reference.
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 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?
Potential energy points along the spring force, so larger $k$ reverses its direction.
$U_{s,1}>U_{s,2}$.
$U_{s,1}<U_{s,2}$ because the larger $k$ means less displacement.
$U_{s,1}=U_{s,2}$ because the stretch is the same.
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.
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?
$U_s(P)=U_s(Q)$.
$U_s(P)<U_s(Q)$ because the spring force is opposite.
$U_s(P)$ is negative while $U_s(Q)$ is positive.
$U_s(P)>U_s(Q)$ because compression stores more energy than stretch.
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.
$U_s(2)>U_s(1)$.
Spring potential energy is a vector, so direction matters.
$U_s(2)=3U_s(1)$ because compression tripled.
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.