Potential Energy - AP Physics 1

Two identical masses are at rest: mass 1 is $3.0\,\text{m}$ above the floor, mass 2 is $1.0\,\text{m}$ above the floor. Floor: $U_g=0$ . Which is true?
Answer: $U_g$ is larger for mass 1 This question tests comparing gravitational potential energies at different heights. For identical masses, gravitational potential energy is Ug = mgh, where h is height above the reference (floor). Mass 1 at 3.0 m has Ug₁ = mg(3.0), while mass 2 at 1.0 m has Ug₂ = mg(1.0). Since 3.0 > 1.0 and m and g are the same for both, Ug₁ > Ug₂. Choice A incorrectly claims mass 2 has more energy because it's closer to the reference, but being closer to the zero reference means less potential energy, not more. Remember that gravitational potential energy increases with height above the reference level.
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?
Answer: $U_g(P)>U_g(R)$ . 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 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?
Answer: $U_s(B)>U_s(A)$ because the stretch magnitude is larger at $B$ . 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.
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?
Answer: $U_g(Y)$ is negative relative to this reference level. 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 puck moves on a frictionless track in Earth's gravitational field. The reference is $U_g=0$ at the floor. Point $X$ is 0.5 m above the floor; point $Y$ is 1.5 m above the floor. Which is correct?
Answer: $U_g(Y)>U_g(X)$ . This question evaluates gravitational potential energy in AP Physics 1. Potential energy is influenced by an object's height above a reference point, such as the floor where $U_g$ = 0. Higher positions result in larger positive $U_g$ = mgh due to greater h. The energy increases with height regardless of the gravitational field's constancy. Choice B mistakenly reverses the comparison, thinking closer to the floor means higher energy, but it's the opposite. Consistently measure heights from the reference to compare $U_g$ across points.
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?
Answer: They are equal at both positions, since $U_s$ depends on $x^2$ . 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.
A spring is compressed $0.10\,\text{m}$ from its relaxed length. Taking $U_s=0$ at $x=0$ , which statement is correct?
Answer: $U_s>0$ . 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.
A pendulum bob is at its lowest point, defined as $U_g=0$ . It is raised to a point $0.30\,\text{m}$ higher. Which statement is correct?
Answer: $U_g$ at the raised point is positive. This question tests understanding of gravitational potential energy in a pendulum system. With $U_g = 0$ at the lowest point, raising the bob $0.30\,\text{m}$ higher gives $U_g = mg(0.30) > 0$ . The raised position has positive gravitational potential energy relative to the reference level. Choice C incorrectly associates the attractive nature of gravity with negative potential energy, but the sign of potential energy depends on position relative to the reference, not on whether the force is attractive. The strategy is to always determine potential energy by comparing the object's position to the chosen reference level, with higher positions having greater potential energy.
A mass hangs from a vertical spring, stretching it $0.06\,\text{m}$ from relaxed length. With $U_s=0$ at zero stretch, which is correct?
Answer: $U_s>0$ . This question tests understanding of elastic potential energy in a vertical spring system. Elastic potential energy is $U_s = \frac{1}{2}kx^2$ , where $x$ is the displacement from the relaxed length. The spring is stretched by $0.06\,\text{m}$ , so $x = 0.06\,\text{m}$ (positive for stretching). Therefore, $U_s = \frac{1}{2}k(0.06)^2 > 0$ . Choice A incorrectly assumes equilibrium means zero potential energy, but equilibrium only means the net force is zero, not that the potential energy is zero. The strategy is to recognize that elastic potential energy depends only on the magnitude of displacement from the relaxed position, not on equilibrium conditions or force directions.
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?
Answer: $U_g(B)>U_g(A)$ because $B$ is above the reference level. 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.
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?
Answer: $U_s(x_2)>U_s(x_1)$ . 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 block is attached to a horizontal spring. At position $A$ the spring is relaxed ( $U_s=0$ ). At position $B$ it is compressed $0.15\,\text{m}$ . Which is true?
Answer: $U_s(B)>U_s(A)$ This question tests understanding of spring potential energy during compression. Spring potential energy is Us = ½kx², where x is the displacement magnitude from the relaxed position. At position A (relaxed), Us(A) = 0. At position B (compressed 0.15 m), Us(B) = ½k(0.15)² > 0. Since any positive value is greater than zero, Us(B) > Us(A). Choice B incorrectly thinks compression makes energy negative, but the x² term ensures potential energy is always positive for any displacement. Remember that spring potential energy is positive whether the spring is compressed or stretched from its relaxed position.
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?
Answer: $U_s(A)>0$ . 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.
A crate is at height $h=1.0\,\text{m}$ where $U_g=0$ is chosen. It is lifted to $h=1.8\,\text{m}$ . What is true about $\Delta U_g$ ?
Answer: $\Delta U_g>0$ . This question tests understanding of changes in gravitational potential energy. The crate moves from $h = 1.0\,\text{m}$ to $h = 1.8\,\text{m}$ , an increase of $0.8\,\text{m}$ . The change in potential energy is $\Delta U_g = mg(1.8) - mg(1.0) = mg(0.8) > 0$ . Since the crate moves upward (to a higher position), its gravitational potential energy increases. Choice A incorrectly claims that the choice of reference level affects the change in potential energy, but while absolute values depend on the reference, changes do not. The key principle is that changes in potential energy depend only on the change in position, not on the choice of reference level.
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?
Answer: The book's $U_g$ is positive relative to the tabletop reference. 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 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?
Answer: $U_g(A)$ is negative. 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.
A block is attached to a horizontal spring and held at the relaxed length $x=0$ , where $U_s=0$ . It is moved to $x=+0.05\,\text{m}$ . Which is correct?
Answer: $U_s>0$ . This question tests understanding of elastic potential energy for positive displacement. Elastic potential energy is $U_s = \frac{1}{2}kx^2$ , where $x$ is displacement from the relaxed position. At $x = +0.05\,\text{m}$ , we have $U_s = \frac{1}{2}k(0.05)^2 > 0$ . The positive value of $x$ indicates stretching (not compression), but the squared term ensures positive energy regardless. Choice A incorrectly confuses zero spring force at the relaxed position with zero force at the displaced position—the spring force is $F = -kx \neq 0$ at $x = 0.05\,\text{m}$ . The strategy is to remember that elastic potential energy is always positive for any non-zero displacement and is a scalar quantity, not a vector.
A block is on a ramp at height $3.0\,\text{m}$ above the ground; $U_g=0$ at the ground. Which statement is correct?
Answer: $U_g>0$ . This question tests understanding of gravitational potential energy on an incline. Gravitational potential energy depends only on vertical height, not on the path or surface. With the block at height $h = 3.0\,\text{m}$ above the ground (where $U_g = 0$ ), we have $U_g = mg(3.0) > 0$ . Choice A incorrectly assumes that support from the ramp affects potential energy, but potential energy depends only on position relative to the reference level, not on supporting forces. The strategy is to remember that gravitational potential energy depends solely on vertical height above the reference level, regardless of the object's support or path.
A spring is horizontal with $U_s=0$ at its natural length. It is compressed by $x=0.10\ \text{m}$ . Compared with $U_s$ at $x=0$ , the spring's potential energy is
Answer: greater 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.
A spring has $U_s=0$ at $x=0$ . It is compressed to $x=-0.12\ \text{m}$ . Compared with $U_s$ at $x=0$ , the spring potential energy is
Answer: greater 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.

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