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FRQ: Spring Force and Newton’s Second Law — The Courier’s Hanging Scale

Assessments aligned to 2026 AP Physics 1 standards

Qualitative/Quantitative Translation (QQT)  |  MID-LEVEL  |  8 points  |  20 min

▤ Scenario

A courier hangs a parcel from a spring scale while riding in an elevator. The top of the scale hangs from a bracket fixed to the elevator’s ceiling, and the parcel, of mass m, hangs from a hook at the scale’s lower end, as shown in Figure 1. Inside the scale is an ideal spring of spring constant k, and a pointer on the spring’s lower end shows the spring’s stretch Δx from its relaxed length on a strip marked in centimeters.

Figure 1 — The courier’s spring scale hanging from a bracket on the elevator’s ceiling.

The elevator moves only vertically. In each case described below, the parcel hangs at rest relative to the elevator, and the parts of the scale that move with the spring’s lower end have negligible mass. Take the ground as an inertial reference frame, with upward as the positive direction.

✎ Free Response Questions

Part A

In Case 1, the elevator moves upward at a constant speed, and the stretch of the spring is Δx_1. In Case 2, the elevator moves downward while its speed decreases at a constant rate, and the stretch of the spring is Δx_2.

Indicate whether Δx_2 is greater than, less than, or equal to Δx_1 by writing one of the following.

•   Δx_2 > Δx_1

•   Δx_2 < Δx_1

•   Δx_2 = Δx_1

Justify your answer using qualitative reasoning beyond referencing equations.

Part B

Consider a general case in which the elevator’s acceleration has vertical component a, where a is positive when the acceleration is directed upward. Starting with Newton’s second law, derive an expression for the stretch Δx of the spring. Express your final answer in terms of m, k, a, 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

Δx_2 > Δx_1

The stretch tells us how hard the spring pulls on the parcel, because an ideal spring’s force is proportional to its stretch. In both cases two forces are exerted on the parcel: the spring’s upward pull and Earth’s downward gravitational force, which is the same in both cases.

In Case 1 the elevator, and the parcel with it, moves with constant velocity. With no change in velocity there is no acceleration, so the two forces balance and the spring pulls up with a force equal to the parcel’s weight. Moving upward does not by itself require any extra force.

In Case 2 the parcel moves downward but slows down, so its change in velocity points upward: its acceleration points up, even though it is moving down. An upward acceleration requires an upward net force, so the spring must now pull up harder than gravity pulls down, and a harder pull requires a greater stretch. The stretch depends on the parcel’s acceleration, not on the direction it is moving.

Figure 2 — Forces on the parcel in each case; arrow lengths compare the forces, not measure them.

Scoring (3 points):

▸ A1 — 1 point: For indicating Δx_2 > Δx_1.

▸ A2 — 1 point: For a justification that indicates that in Case 1 the parcel’s velocity is constant, so the forces on it are balanced and the spring’s force equals the gravitational force on the parcel.

▸ A3 — 1 point: For a justification that indicates that in Case 2 the parcel’s acceleration is directed upward, so the spring’s upward force must exceed the gravitational force on the parcel, which requires a greater stretch.

Scoring Note: A2 and A3 are content criteria: any wording with that content satisfies them, and the words “net force” are not required. They are judged on the reasoning given; an equation with no accompanying reasoning satisfies neither. A2 and A3 may each be earned independently of the option indicated in A1.

⚠︎ Common error (partial credit): Indicates Δx_2 > Δx_1 and argues correctly that the upward acceleration in Case 2 requires the spring to pull harder than gravity, and so to stretch more, but never explains why the pull in Case 1 equals the weight — earns 2 of 3 points, A1 and A3.

⚠︎ Common error (partial credit): Indicates Δx_2 > Δx_1 only because the elevator is slowing down, so the parcel “pulls down harder”, without identifying an upward acceleration or comparing the spring’s force with the weight in either case — earns 1 of 3 points, A1.

⚠︎ Common error (partial credit): Indicates Δx_2 = Δx_1, arguing correctly that constant velocity in Case 1 means the spring’s pull equals the weight, and assuming the same holds in Case 2 because the parcel’s weight has not changed — earns 1 of 3 points, A2. In Case 2 the velocity is changing, so the forces are not balanced.

✗ Common error (no credit): Indicates Δx_2 < Δx_1 because the parcel moves downward in Case 2 and upward in Case 1, so it “feels lighter” going down, with no force reasoning about either case — earns 0 points. The stretch depends on the direction of the parcel’s acceleration, not the direction of its velocity, and slowing down while moving down is an upward acceleration.

Part B — Model Answer

Take upward as positive. Two forces are exerted on the parcel: the spring’s upward pull and Earth’s downward gravitational force mg. With upward positive, the spring’s lower end is displaced by −Δx from its relaxed position, so F_s = −k(−Δx) = +kΔx: the two minus signs cancel, and the spring pulls the parcel upward with a force of magnitude kΔx. The parcel hangs at rest relative to the elevator, so it shares the elevator’s acceleration a:

ΣF = ma

kΔx − mg = ma

Δx = m(g + a)/k

With a = 0 this gives mg/k, the stretch of a spring holding a mass at rest. The spring’s pull differs from the gravitational force on the parcel by exactly the net force, ma.

Scoring (3 points):

▸ B1 — 1 point: For a multistep derivation that includes Newton’s second law. A final expression for Δx on its own is not a multistep derivation.

▸ B2 — 1 point: For indicating that the spring exerts a force of magnitude kΔx on the parcel (Hooke’s law), directed upward.

▸ B3 — 1 point: For a correct expression, Δx = m(g + a)/k, consistent with the indicated spring force. Where B2 was not earned, an expression that follows correctly from the response’s own spring-force expression also earns this point; any other error prevents it.

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

Scoring Note: B1 and B2 do not depend on any other part, and neither carries a consistency clause.

⚠︎ Common error (partial credit): Substitutes Hooke’s law as written, F_s = −kΔx, directly into ΣF = ma with upward positive, obtaining −kΔx − mg = ma and Δx = −m(g + a)/k — earns 2 of 3 points, B1 and B3. With Δx the size of the stretch, the spring’s lower end is displaced by −Δx, so the spring’s force is +kΔx, upward.

⚠︎ Common error (partial credit): Starts from ΣF = ma but sets a = 0 because the parcel is at rest relative to the elevator, obtaining Δx = mg/k — earns 2 of 3 points, B1 and B2. The parcel is at rest relative to the elevator, not the ground, so it shares the elevator’s acceleration.

⚠︎ Common error (partial credit): Takes downward as positive for the forces, writing mg − kΔx = ma, but keeps a as the upward component defined in the prompt, and obtains Δx = m(g − a)/k — earns 2 of 3 points, B1 and B2. With downward positive, an upward acceleration enters as −a.

⚠︎ Common error (partial credit): Leaves out the gravitational force, writing kΔx = ma, and obtains Δx = ma/k — earns 2 of 3 points, B1 and B2. Earth pulls on the parcel throughout; without the spring it would accelerate downward at g.

⚠︎ Common error (partial credit): Writes the force balance kΔx = mg directly, with no Newton’s second law, and obtains Δx = mg/k — earns 1 of 3 points, B2. Unless a = 0, the forces on the parcel are unbalanced.

Part C — Model Answer

The derived equation is consistent with the reasoning in part A.

Part A’s reasoning was that the spring’s pull, and so its stretch, depends on the parcel’s acceleration and not on its velocity. The equation Δx = m(g + a)/k says exactly that: the stretch increases linearly with a, and neither the elevator’s speed nor its direction of motion appears in it.

In Case 1 the velocity is constant, so a = 0 and Δx_1 = mg/k. In Case 2 the elevator moves downward and slows, so its acceleration is upward, a > 0, and Δx_2 = m(g + a)/k, which is greater than mg/k. The equation therefore gives Δx_2 > Δx_1, as argued in part A. Velocity does not appear in the equation, so of the elevator’s motion only its acceleration affects the stretch.

Scoring (2 points):

▸ C1 — 1 point: For attempting to address the functional dependence between Δx and a, or between Δx and the elevator’s velocity, in the equation derived in part B.

Scoring Note: C1 does not require the functional dependence to be used correctly. Functional-dependence language, such as “increases with”, “increases linearly with”, “depends on” or “does not appear in”, relating Δx, as it appears in the response’s own part B expression, to a or to the velocity, is sufficient.

▸ C2 — 1 point: For correctly using that functional dependence to decide whether the part B equation agrees with part A, judged against the response’s own part B expression and its own part A answer, and stating the verdict that follows — consistent or not consistent.

Scoring Note: C2 requires the correct acceleration in each case: a = 0 in Case 1 and a > 0 in Case 2. An argument from velocity alone satisfies C1 but not C2. The verdict may be given in any words.

⚠︎ Common error (partial credit): Notes that Δx increases with a in the part B equation but gives no verdict on whether the equation agrees with part A — earns 1 of 2 points, C1.

⚠︎ Common error (partial credit): Takes the acceleration in Case 2 to be downward because the elevator moves downward, so that a < 0 makes Δx_2 smaller, and gives whichever verdict follows from its own part A answer — earns 1 of 2 points, C1. Slowing down while moving downward is an upward acceleration.

✗ Common error (no credit): Asserts that the equation agrees with part A with no statement of how Δx depends on a or on the velocity — earns 0 points. A verdict with no functional-dependence reasoning is a claim, not a justification.

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