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FRQ: Work Done by a Lifting Force — The Stockroom Shelf Lifts

Assessments aligned to 2026 AP Physics 1 standards

Qualitative/Quantitative Translation (QQT)  |  CHALLENGE  |  8 points  |  20 min

▤ Scenario

Two identical crates, each of mass m, are raised from rest on a stockroom floor to rest on a shelf a height h above the floor. Crate 1 is raised by a forklift, which lifts it slowly and smoothly. Crate 2 is raised by a cable hoist, which lifts it quickly: the cable tension is greater than mg while the crate speeds up and less than mg while it slows, and the crate arrives at the shelf in much less time than Crate 1. Figure 1 shows the two lifts. Let W_1 be the work done on Crate 1 by the forklift during its lift, and W_2 the work done on Crate 2 by the cable during its lift.

Figure 1. Crate 1 on the forklift and Crate 2 on the hoist, each raised from rest on the floor to rest on the shelf at height h.

Air resistance is negligible, and each crate moves vertically without swinging or tipping. Use the floor as the reference frame.

✎ Free Response Questions

Part A

Indicate whether W_2 is greater than, less than, or equal to W_1 by writing one of the following.

• W_2 > W_1

• W_2 < W_1

• W_2 = W_1

Justify your answer using qualitative reasoning beyond referencing equations.

Part B

A third crate, also of mass m, is raised from rest on the floor by a cable and passes the level of the shelf, a height h above the floor, moving upward with speed v. Starting with the work–energy theorem, derive an expression for the work W done on this crate by the cable during the lift. Express your final answer in terms of m, h, v, and physical constants, as appropriate. Begin your derivation by writing the fundamental physics principle or an equation from the reference information.

Part C

Justify how your derived equation in part B is or is not consistent with your reasoning in part A.

❖ 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 — Model Answer

W_2 = W_1. Both crates are at rest at the start and at rest at the end, so neither crate has any change in kinetic energy: over the whole lift, the energy the lifting force puts into a crate must exactly equal the energy the gravitational force takes out of it. The gravitational force is the same on both crates and each is raised through the same vertical displacement h, so gravity removes the same amount of energy from each. The lifting force must therefore transfer the same energy to each crate, whatever the size of the force at any instant and however long the lift takes. The hoist pulls harder than mg at first — but it also pulls less hard than mg later, and with no kinetic energy left over at the top the two effects cancel exactly.

Scoring (3 points):

▸ A1 — 1 point: For indicating W_2 = W_1.

▸ A2 — 1 point: For reasoning that each crate is at rest at both ends of its lift, so there is no change in its kinetic energy — the energy transferred in by the lifting force equals the energy transferred out by gravity (or, equivalently, the lifting force's work equals the increase in the crate–Earth system's gravitational potential energy).

▸ A3 — 1 point: For reasoning that the energy gravity removes (or the potential energy gained) depends only on the crate's weight and the height h, which are the same for both crates, so the lifting force must transfer the same energy to each regardless of the force at any instant or the time taken.

⚠︎ Common error (partial credit): Selects W_2 = W_1 with the justification "both crates gain the same gravitational potential energy mgh, whatever the force" and no reference to the crates being at rest at both ends — earns A1 and A3 (2 of 3 points).

✗ Common error (no credit): Selects W_2 > W_1 because the hoist exerts a larger force — earns 0 of 3 points: the larger tension acts only while the crate speeds up, and a smaller tension while it slows.

✗ Common error (no credit): Selects W_2 < W_1 because the hoist takes less time — earns 0 of 3 points: it confuses work with power.

Part B — Model Answer

Treat the crate as the object. Two forces do work on it during the lift: the cable tension (upward, along the displacement — positive work W) and the gravitational force (downward, opposite the displacement h — negative work). The crate starts from rest and arrives moving at speed v.

W_net = ΔK = K_f − K_i

W + W_g = ½mv² − 0

W + (−mgh) = ½mv²

W = mgh + ½mv²

The cable must supply the energy gravity removes, mgh, plus whatever kinetic energy the crate has on arrival. A crate that arrives at rest (v = 0) needs exactly mgh of work from the cable; one that arrives moving needs more.

Scoring (3 points):

▸ B1 — 1 point: For a multistep derivation that includes the work–energy theorem, with the net work written as the sum of the work done by the cable and the work done by the gravitational force.

▸ B2 — 1 point: For the work done by the gravitational force expressed as −mgh (negative) and the change in kinetic energy as ½mv² − 0.

▸ B3 — 1 point: For the correct final expression W = mgh + ½mv².

Scoring Note: A correct, isolated, final expression for W earns points B2 and B3.

⚠︎ Common error (partial credit): Sets the change in kinetic energy to zero and obtains W = mgh, overlooking that the crate arrives at speed v — earns B1 (1 of 3 points).

⚠︎ Common error (partial credit): Writes the work done by gravity as +mgh and obtains W = ½mv² − mgh — earns B1 only (1 of 3 points).

✗ Common error (no credit): Writes W = Fh with F = mg, with no principle stated — earns 0 of 3 points: it asserts a constant lifting force and never uses the change in kinetic energy.

Part C — Model Answer

Consistent. For the two crates in part A, each arrives at the shelf at rest, so v = 0 and the expression from part B reduces to W = mgh for both: W_1 = W_2 = mgh. The expression contains no quantity describing how the lift was carried out — no tension, no acceleration, no time — only the weight of the crate, the height gained, and the kinetic energy it has on arrival. That is exactly the part A reasoning: with no change in kinetic energy, the lifting force's work is fixed by what gravity removes, mgh, whatever the force at any instant. The ½mv² term is the only way the manner of the lift can enter, and it is zero whenever the crate ends at rest.

Scoring (2 points):

▸ C1 — 1 point: For substituting v = 0 into the part B expression (or otherwise applying it to crates that arrive at rest) to obtain W = mgh for both crates, and concluding that this agrees with W_2 = W_1 from part A. This point may be earned with an expression consistent with the student's own answer to part B.

▸ C2 — 1 point: For explaining that the derived expression depends only on m, h and the arrival speed v — not on the size of the lifting force or the time taken — so a larger or briefer lifting force cannot change W unless it leaves the crate with kinetic energy, matching the qualitative reasoning in part A.

⚠︎ Common error (partial credit): Writes W = mgh for both crates and says "consistent", without saying why the ½mv² term is absent or why the manner of the lift drops out — earns C1 (1 of 2 points).

✗ Common error (no credit): Claims the two parts are inconsistent because the hoist's larger tension must appear in the expression — earns 0 of 2 points: the expression shows that only the arrival kinetic energy, not the tension, distinguishes one lift from another.

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