Pause here: To build true exam stamina, attempt these questions on your own before checking the model answers.

FRQ: Gravitational Potential Energy of a System — The Harbourview Drop Tower

Assessments aligned to 2026 AP Physics 1 standards

Mathematical Routines (MR)  |  CHALLENGE  |  10 points  |  25 min

▤ Scenario

The Harbourview amusement park runs a drop-tower ride. A passenger gondola of total mass m = 1200 kg (passengers included) is winched to the top of a vertical tower and held at rest. When it is released, the gondola falls freely for a distance D = 36 m before magnetic brakes engage, as shown in Figure 1. Let d be the distance the gondola has fallen since release, so d = 0 at the release point and d = D at the instant the brakes engage.

Take the gondola–Earth system's gravitational potential energy to be zero at the point where the brakes engage (d = D). Air resistance is negligible during the free fall, and the gondola's motion is vertical. Use the ground as the reference frame and take upward as the positive y direction. Use g = 10 m/s².

Figure 1: The gondola at release (d = 0) and, dashed, at the brake point (d = D) a distance D below.

✎ Free Response Questions

The following information applies to parts A and B.

Part A

i. On the axes provided in Figure 2, sketch the gravitational potential energy U_g of the gondola–Earth system as a function of the distance fallen d, from d = 0 to d = D.

Figure 2: Axes for part A (i) — U_g of the gondola–Earth system against the distance fallen d.

ii. Derive an expression for the change in gravitational potential energy ΔU_g of the gondola–Earth system as the gondola falls from d = 0 to d = D. Express your final answer in terms of m, g, D, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

iii. Derive an expression for the speed v of the gondola at the instant it has fallen a distance d. Express your final answer in terms of g, d, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information. Then calculate the speed of the gondola at the instant the brakes engage.

Part B

A student watching the ride makes the following claim: "During the fall, the gravitational force does positive work mgD on the gondola, and the gondola–Earth system also loses potential energy mgD. So the gondola's kinetic energy at the brake point must be 2mgD."

Indicate whether the student's claim is correct or incorrect by writing one of the following.

______ Correct

______ Incorrect

Briefly justify your response by referencing a fundamental physics principle.

❖ Answer Key & Scoring Guide

▸ earns credit  ⚠︎ common error, partial credit  ✗ common error, no credit

Points are independent and there are no deductions. ▸ criteria follow College Board's published scoring standard; ⚠︎ and ✗ lines are TheScienceCube teaching commentary, not College Board rubric.

Part A (i) — Model Answer

With U_g = 0 at the brake point, the system's potential energy at a distance d fallen is U_g = mg(D − d): it is mgD at release and falls in a straight line to zero at d = D. The graph is a straight line because near the surface U_g changes in proportion to the vertical position, with slope of magnitude mg.

Part A (i) answer: U_g = mg(D − d), a straight line from mgD at d = 0 to zero at d = D.

Scoring (2 points):

▸ A1 — 1 point: For a single straight line with negative slope over the whole interval 0 ≤ d ≤ D.

▸ A2 — 1 point: For a line that starts at a positive value at d = 0 and reaches U_g = 0 exactly at d = D (consistent with the stated zero of potential energy).

⚠︎ Common error (partial credit): A straight line of negative slope that ends above the axis at d = D, or below it before d = D — earns A1 of the 2 points; the zero was placed at the brake point by the question.

✗ Common error (no credit): A curve that bends (concave up or down) because "the gondola speeds up" — the speed is irrelevant to U_g, which depends only on position; this confuses the kinetic-energy graph with the potential-energy graph.

Part A (ii) — Model Answer

Begin with the reference-information result for a system near a planet's surface:

ΔU_g = mgΔy

The gondola's vertical position changes by Δy = −D (it moves downward a distance D), so

ΔU_g = mg(−D) = −mgD

The gondola–Earth system's gravitational potential energy decreases by mgD. The sign is negative because Δy is negative: the object moved down, and potential energy is lost from the system.

Scoring (2 points):

▸ A3 — 1 point: For starting from ΔU_g = mgΔy, or from U_g = mgy evaluated at the two positions, before any substitution.

▸ A4 — 1 point: For the final expression ΔU_g = −mgD, with the negative sign.

Scoring Note: A correct, isolated, final expression for ΔU_g earns point A4.

⚠︎ Common error (partial credit): ΔU_g = +mgD from a correct start but with Δy taken as +D — earns A3 of the 2 points; the direction of Δy was dropped.

✗ Common error (no credit): Writing the change in potential energy as the work done by gravity, +mgD, with no reference to Δy or to U_g — the work done by gravity on the gondola is a different bookkeeping of the same energy (see part B), and the question asks for ΔU_g of the system.

Part A (iii) — Model Answer

Treat the gondola alone as the system. Only the gravitational force does work on it, so the work–energy theorem gives

W_net = ΔK → W_g = ½mv² − 0

The gravitational force mg acts downward and the point of application moves downward a distance d, so W_g = +mgd. Then

mgd = ½mv² → v = √(2gd)

At the brake point, d = D = 36 m:

v = √(2 × 10 m/s² × 36 m) = √720 m/s ≈ 27 m/s

The mass cancels: every gondola, loaded or empty, reaches the brake point at the same speed.

Scoring (3 points):

▸ A5 — 1 point: For a derivation that begins with the work–energy theorem (W_net = ΔK, or an equivalent statement that the work done by gravity equals the change in kinetic energy), with the initial kinetic energy zero.

▸ A6 — 1 point: For the final expression v = √(2gd), with the mass cancelled.

▸ A7 — 1 point: For a calculated speed of 27 m/s (accept 26.8 m/s to 27 m/s), consistent with the expression in A6.

Scoring Note: A correct, isolated, final expression for v earns point A6. Point A7 may be earned for a value consistent with an incorrect expression in A6.

⚠︎ Common error (partial credit): Writing W_g = −mgd on the fall and arriving at an imaginary or sign-flipped result — earns A5 of the 3 points; on the way down the force and the displacement point the same way, so the work is positive.

⚠︎ Common error (partial credit): v = 2gd, or v = √(gd), from an algebra slip after a correct start — earns A5 and a consistent A7.

✗ Common error (no credit): Starting from W_g = mgD and ΔU_g = −mgD added together to give ½mv² = 2mgD — the double count that part B asks the student to reject; this earns no point in A (iii) because the starting principle is misapplied.

Part B — Model Answer

The claim is incorrect. The work done by gravity and the decrease in the system's potential energy are two ways of describing the same energy transfer, and only one of them may appear in any one energy equation.

With the gondola alone as the system, gravity is an external force. It does work W_g = +mgD on the gondola, and by the work–energy theorem that work is the whole change in kinetic energy: K = mgD. In this picture there is no potential energy at all, because a system consisting of a single object can only have kinetic energy.

With the gondola and the Earth as the system, gravity is an internal interaction. Its effect is recorded as the change in the system's potential energy, ΔU_g = −mgD (part A (ii)), which is balanced by an equal increase in kinetic energy, K = mgD. In this picture gravity does not appear as a separate work term, because no energy crosses the system boundary. Either bookkeeping gives K = mgD; adding them counts the same energy twice.

Scoring (3 points):

▸ B1 — 1 point: For indicating that the claim is incorrect.

▸ B2 — 1 point: For stating that the work done by gravity on the gondola and the decrease in the gondola–Earth system's potential energy describe the same energy transfer, so only one may be used — or, equivalently, that with the Earth outside the system there is no potential energy, and with the Earth inside the system gravity does no work across the boundary.

▸ B3 — 1 point: For concluding, from the work–energy theorem or from the balance of kinetic and potential energy within the system, that the kinetic energy at the brake point is mgD, not 2mgD.

Scoring Note: Point B2 may be earned by a response that names the system explicitly for either bookkeeping; both are not required.

⚠︎ Common error (partial credit): "Incorrect, because energy cannot be created" with no reference to the choice of system or to the two bookkeepings — earns B1 of the 3 points; the justification must say why the two terms are not additive.

⚠︎ Common error (partial credit): "Incorrect", followed by a correct calculation K = ½mv² = mgD from part A (iii) but no statement about work and potential energy being alternatives — earns B1 and B3.

✗ Common error (no credit): "Correct — gravity does work and the potential energy also drops, so both add" — the double count: it treats the gravitational force as external to the system while also assigning that system a gravitational potential energy.

Why Can't You Add Gravity's Work to the Potential Energy Lost?

In short: The work gravity does on the gondola and the drop in the gondola–Earth system's potential energy are two ways of recording the same energy transfer. With the gondola alone as the system, gravity does +mgD of work and K = mgD. With the gondola and Earth as the system, potential energy falls by mgD and K rises by mgD. Use one account, never both, so the kinetic energy at the brake point is mgD, not 2mgD.

Common doubts

Why is ΔU negative when gravity does positive work? ΔU = mgΔy, and Δy = −D because the gondola moves down, so ΔU = −mgD. Gravity's work on the gondola is +mgD because the force and the displacement both point down. It is the same energy, recorded with two different signs.

Does the gondola's mass change its speed at the brake point? No. Setting mgd = ½mv² cancels the mass, so v = √(2gd). With D = 36 m and g = 10 m/s², every gondola, loaded or empty, reaches the brakes at about 27 m/s.

Why is the U–d graph a straight line when the gondola speeds up? Potential energy depends only on position, not on speed. With U = 0 at the brake point, U = mg(D − d), a straight line of slope −mg. The speeding up shows in the kinetic energy graph, not this one.

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