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FRQ: Work Done by Individual Forces — The Rapid Crane Lift

Assessments aligned to 2026 AP Physics 1 standards

Translation Between Representations (TBR)  |  MID-LEVEL  |  12 points  |  30 min

▤ Scenario

A tower crane lifts a steel beam of mass m = 500 kg vertically upward from rest at ground level, as shown in Figure 1. From y = 0 to y = h_1 = 4.0 m the winch is set so that the cable exerts on the beam a constant upward force of magnitude T = 6000 N. The cable remains vertical, the beam does not swing or rotate, and air resistance is negligible. The beam leaves the ground the instant the lift begins and thereafter touches nothing but the cable.

Figure 1 — The beam at the start of the lift; the first part of the lift ends at y = h_1 = 4.0 m.

Take the ground as an inertial reference frame, with y measured upward from the beam’s starting position. All motion is one-dimensional. Use g = 10 N/kg.

✎ Free Response Questions

Part A

On the dot in Figure 2, which represents the beam, draw and label the forces (not components) that are exerted on the beam after it has left the ground, while it is between y = 0 and y = h_1.

•   Each arrow must start on, and point away from, the dot.

•   Draw the length of each arrow to represent the magnitude of each force, consistent with the scale used in Figure 2.

Figure 2 — The dot represents the beam. Each square of the grid represents 1000 N.

Part B

Starting with the work–energy theorem or Newton’s second law, derive an expression for the speed v_1 of the beam when it reaches y = h_1. Express your final answer in terms of m, T, h_1, and physical constants, as appropriate. Begin your derivation by writing the fundamental physics principle or an equation from the reference information.

Part C

Figure 3 shows a graph of energy as a function of the height y of the beam, from y = 0 to y = h_1. The work W_T done on the beam by the cable is shown.

On the axes shown in Figure 3, do the following.

i.    Sketch and label a line or curve that represents the work W_g done on the beam by the gravitational force, as a function of y from y = 0 to y = h_1.

ii.   Sketch and label a line or curve that represents the kinetic energy K of the beam, as a function of y from y = 0 to y = h_1.

Figure 3 — Work done on the beam by the cable, W_T, against the beam’s height y, for part C.

Part D

After the beam reaches y = h_1, the operator reduces the tension in the cable so that the beam continues upward at constant velocity until it reaches y = h_2 = 2h_1.

Predict how the line representing W_T in Figure 3 and the line representing K that you drew in part C would each continue for heights between h_1 and h_2. Justify your prediction using consistent reasoning.

❖ 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

Two forces are exerted on the beam: the cable pulls it upward with the tension T = 6000 N, and Earth pulls it downward with the gravitational force F_g = mg = (500 kg)(10 N/kg) = 5000 N. On the 1000 N grid the tension arrow is six squares long and the gravitational arrow five squares long, so the upward arrow is longer.

Figure 4 — Forces on the beam during the first part of the lift, drawn to the scale of Figure 2.

Nothing else touches the beam, so there is no third force. The beam’s velocity and acceleration are not forces and get no arrows.

Scoring (3 points):

▸ A1 — 1 point: For drawing exactly two arrows starting on the dot: one directed upward and labelled as the tension or cable force, and one directed downward and labelled as the gravitational force or weight, with no other force.

▸ A2 — 1 point: For drawing the upward arrow six squares long, consistent with T = 6000 N on the scale of Figure 2.

▸ A3 — 1 point: For drawing the downward arrow five squares long, consistent with mg = 5000 N on the scale of Figure 2.

Scoring Note: Any label that names the force is accepted — T, F_T, tension or cable force; F_g, mg, W or weight. The response is not required to name the object exerting each force. A2 and A3 are each scored on that arrow alone and may be earned with or without A1.

⚠︎ Common error (partial credit): Draws the two forces with correct lengths but adds an upward arrow labelled as the force of motion, F_net, or ma — earns 2 of 3 points, A2 and A3. The net force is the resultant of the forces drawn, not a third force, and motion itself is not a force.

⚠︎ Common error (partial credit): Draws the two forces correctly directed and labelled but with equal lengths of five squares, as if the forces were balanced — earns 2 of 3 points, A1 and A3. The beam accelerates upward, so the tension exceeds the gravitational force; the given T = 6000 N is six squares.

✗ Common error (no credit): Draws a single upward arrow labelled F_net or F = 1000 N, one square long — earns 0 points. A free-body diagram shows the individual forces exerted on the object, not their resultant.

Part B — Model Answer

Apply the work–energy theorem to the beam from y = 0, where it is at rest, to y = h_1. The cable force acts along the upward displacement and does positive work Th_1; the gravitational force acts against it and does negative work −mgh_1. These are the only forces.

ΔK = W_net = W_T + W_g
½mv_1² − 0 = Th_1 − mgh_1
v_1 = √[2(T − mg)h_1 / m]

For this lift, v_1 = √[2(1000 N)(4.0 m)/(500 kg)] = √16 m/s = 4.0 m/s. The cable does 24 kJ of work on the beam and gravity does −20 kJ. Only the 4 kJ difference between the two works becomes kinetic energy.

Scoring (4 points):

▸ B1 — 1 point: For a multistep derivation that includes the work–energy theorem (ΔK = W_net) or Newton’s second law (T − mg = ma) together with a constant-acceleration equation. This point is earned for the correct starting principle alone, even if the steps that follow are incorrect, but at least one step must follow.

▸ B2 — 1 point: For indicating that the cable does work +Th_1 and the gravitational force does work −mgh_1 on the beam — or, on the Newton’s-law route, that the net force is T − mg.

▸ B3 — 1 point: For setting the net work (T − mg)h_1 equal to the change in kinetic energy ½mv_1² with the initial kinetic energy zero — or, on the Newton’s-law route, for substituting a = (T − mg)/m into v_1² = 2ah_1.

▸ B4 — 1 point: For a correct expression, v_1 = √[2(T − mg)h_1/m], or equivalent.

Scoring Note: A correct, isolated, final expression for v_1 earns points B2, B3 and B4.

⚠︎ Common error (partial credit): Sets ½mv_1² = Th_1, omitting the work done by the gravitational force, and obtains v_1 = √(2Th_1/m) — earns 1 of 4 points, B1. The beam is lifted against gravity, which does negative work over the same displacement.

⚠︎ Common error (partial credit): Writes the net work as (T + mg)h_1 and obtains v_1 = √[2(T + mg)h_1/m] — earns 1 of 4 points, B1. The gravitational force is directed against the upward displacement, so its work is negative.

⚠︎ Common error (partial credit): Writes ΔK = (T − mg)h_1 and then equates it to mv_1 instead of ½mv_1² — earns 2 of 4 points, B1 and B2; the net work was never set equal to ½mv_1², and the final expression is incorrect.

✗ Common error (no credit): Writes v_1 = √(2gh_1), treating the beam as if it were in free fall, with no principle stated — earns 0 points. The beam is not falling and its acceleration is not g; no work or force statement was made.

Part C (i) — Model Answer

The gravitational force is constant and directed against the upward displacement, so its work is W_g = −mgy: a straight line through the origin with negative slope, reaching −(5000 N)(4.0 m) = −20 kJ at y = h_1. Its slope is smaller in magnitude than that of the W_T line, because mg < T.

Part C (ii) — Model Answer

The beam starts from rest, so its kinetic energy at any height equals the net work done on it so far: K = W_T + W_g = (T − mg)y. That is a straight line through the origin with slope T − mg = 1000 N, reaching (1000 N)(4.0 m) = 4 kJ at y = h_1 — the vertical gap between the W_T line and the W_g line at every height.

Figure 5 — The lines for W_g and K added to Figure 3; K is the sum of the two work lines.

At every height the K line sits exactly as far above the axis as W_T is above |W_g|. Kinetic energy tracks the sum of the works, not the work of the cable alone.

Scoring (3 points):

▸ C1 — 1 point: For sketching a straight line, decreasing from the origin and continuous from y = 0 to y = h_1, labelled W_g or gravitational work.

▸ C2 — 1 point: For the response’s W_g line, drawn from the origin, passing through (4.0 m, −20 kJ) within half a gridline, so that its magnitude of slope is smaller than that of W_T.

▸ C3 — 1 point: For sketching a straight line from the origin, labelled K, whose value at y = h_1 equals W_T plus the response’s own W_g value, within half a gridline — +4 kJ for a correct W_g line — or, where that sum lies above the axes, a straight line from the origin steeper than W_T that runs off the top of the grid.

⚠︎ Common error (partial credit): Draws W_g correctly but draws K as a line coinciding with W_T, as if all of the cable’s work became kinetic energy — earns 2 of 3 points, C1 and C2. The gravitational force removes energy over the same displacement.

⚠︎ Common error (partial credit): Draws W_g as a straight line from the origin to (4.0 m, −24 kJ), the mirror image of W_T, and draws K along the horizontal axis, as though the two works cancelled because the forces “balance” — earns 2 of 3 points, C1 and C3: the K line is consistent with the response’s own W_g line, and the error is in W_g itself. The tension exceeds mg, so the works do not cancel; the gap between the two lines is the kinetic energy.

⚠︎ Common error (partial credit): Draws W_g as a curve bending downward, reaching −20 kJ at h_1, and K correctly — earns 2 of 3 points, C2 and C3. The gravitational force is constant, so its work grows in proportion to y.

✗ Common error (no credit): Draws W_g as a horizontal line at −20 kJ and K as a horizontal line at +4 kJ, plotting the final values as constants — earns 0 points. Neither line is drawn from the origin; work and kinetic energy accumulate with height, and neither is constant during the lift.

Part D — Model Answer

At constant velocity the beam’s acceleration is zero, so by Newton’s second law the net force on it is zero: the tension is reduced to T′ = mg = 5000 N. Between h_1 and h_2 the cable therefore does work mg·Δy and the gravitational force does work −mg·Δy; the net work is zero, and by the work–energy theorem the kinetic energy does not change.

On Figure 3, the W_T line continues to rise but with a smaller slope — 5000 N instead of 6000 N, equal in magnitude to the slope of the W_g line — while the K line becomes horizontal at 4 kJ from h_1 to h_2.

Figure 6 — The three lines continued to h_2 at constant velocity: K is flat, W_T and W_g cancel.

The cable is still doing positive work on the beam the whole way up, and gravity is still doing negative work; they now cancel exactly. Constant speed does not mean no work is being done — it means the net work is zero.

Scoring (2 points):

▸ D1 — 1 point: For predicting that the W_T line continues to increase with a smaller slope than before and that the K line is horizontal (constant) between h_1 and h_2.

▸ D2 — 1 point: For a justification that relates constant velocity to zero net force (T = mg) or zero net work, so that the cable’s positive work and the gravitational force’s negative work cancel and the kinetic energy does not change.

Scoring Note: D1 does not require the response to state that the new slope of W_T equals the magnitude of the slope of W_g. D2 is earned for relating constant velocity to zero net force or to zero net work; the cancellation of the two works need not be stated. D1 may be earned with or without D2.

⚠︎ Common error (partial credit): Predicts a horizontal K line and justifies it from zero net force, but predicts that the W_T line also becomes horizontal because “the tension no longer does work” — earns 1 of 2 points, D2. The cable still exerts an upward force on a beam that is moving upward; it does positive work, now equal in magnitude to the gravitational force’s negative work.

⚠︎ Common error (partial credit): Predicts both lines correctly but justifies the flat K line only by saying that the speed is constant, so ½mv² is constant, with no reference to net force or net work — earns 1 of 2 points, D1. Restating the definition of K is not a justification; D2 is earned for relating constant velocity to zero net force or zero net work.

✗ Common error (no credit): Predicts that the K line continues to rise with the same slope because the beam is still moving upward and the cable is still pulling — earns 0 points. This confuses motion with net force: at constant velocity the net work, and hence the change in kinetic energy, is zero.

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