United StatesPhysics C: MechanicsSyllabus dot point

How does an ideal spring exert a restoring force proportional to its displacement, and how does this linear force law set up oscillation and elastic energy?

Topic 2.8 Spring Forces: model the ideal spring with Hooke's law as a linear restoring force, combine springs in series and parallel, and connect the force law to elastic potential energy by integration.

A focused answer to AP Physics C: Mechanics Topic 2.8, covering the ideal spring and Hooke's law as a linear restoring force, the sign convention for the restoring direction, effective spring constants for series and parallel combinations, and the link to elastic potential energy by integrating the force, with worked examples.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-04

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What this topic is asking
Hooke's law and the restoring force
Combining springs
From force to elastic potential energy
Try this

What this topic is asking

The College Board (Topic 2.8) wants you to model an ideal spring with Hooke's law, a restoring force proportional to displacement, to handle springs combined in series and parallel, and to connect the linear force law to elastic potential energy by integration. Springs reappear throughout the course: they set up simple harmonic motion in Unit 7 and store elastic energy in Unit 3, both of which follow directly from the force law established here.

Hooke's law and the restoring force

The defining property is that the force is a linear restoring force: it is proportional to the displacement and always points back toward equilibrium, which is what the minus sign encodes. Stretch the spring and it pulls back; compress it and it pushes out. A stiffer spring (larger $k$ ) exerts more force for the same displacement. This single linear law is what makes a mass on a spring oscillate in simple harmonic motion, the topic that opens Unit 7.

Combining springs

When springs act together, an effective spring constant describes the combination. In parallel (side by side, sharing the displacement) the forces add, so the effective constant is the sum:

k_{eq} = k_1 + k_2 \quad \text{(parallel)}.

In series (end to end, sharing the force but adding extensions) the reciprocals add:

\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} \quad \text{(series)}.

Parallel springs are stiffer than either alone; series springs are softer. Notice these rules are the opposite of the resistor combination rules, which sometimes catches students out.

From force to elastic potential energy

Because the spring force varies with position, the work it does (and the energy it stores) comes from integrating the force over the displacement. The energy stored when a spring is stretched from $0$ to $x$ is

U = \int_0^x F\,dx' = \int_0^x kx'\,dx' = \tfrac{1}{2}kx^2.

This elastic potential energy scales with the square of the displacement, so doubling the stretch quadruples the stored energy, even though it only doubles the force. Graphically, $\tfrac{1}{2}kx^2$ is the triangular area under the linear $F = kx$ graph. This integral is the prototype for the general definition of work done by a variable force, and it feeds straight into the energy conservation of Unit 3.

A vertical spring with a hanging mass

A spring with $k = 120$ N/m hangs vertically. A $0.60$ kg mass is attached and allowed to come to rest. Take $g = 9.8$ m/s squared. Find the equilibrium extension and the elastic energy stored.

step 1 Set up the equilibrium condition

At rest the upward spring force balances the weight: $kx = mg$ .

step 2 Solve for the extension

$x = \dfrac{mg}{k} = \dfrac{(0.60)(9.8)}{120} = \dfrac{5.88}{120} = 0.049$ m, about $4.9$ cm.

step 3 Find the stored elastic energy

$U = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(120)(0.049)^2 = \tfrac{1}{2}(120)(0.0024) = 0.14$ J.

step 4 Interpret

The mass settles where the linearly growing spring force first equals its weight; the energy stored there is the work the spring did resisting the stretch.

Try this

Q1. Two springs of $k = 100$ N/m each are connected in parallel. Calculate the effective spring constant. [2 points]

Cue. Parallel springs add: $k_{eq} = 100 + 100 = 200$ N/m.

Q2. A spring stores $2.0$ J of elastic energy at an extension of $0.10$ m. Calculate its spring constant. [2 points]

Cue. $U = \tfrac{1}{2}kx^2 \Rightarrow 2.0 = \tfrac{1}{2}k(0.10)^2 \Rightarrow k = \dfrac{4.0}{0.01} = 400$ N/m.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (short FRQ). A spring obeys Hooke's law with force constant

k = 250

N/m. (a) Calculate the force needed to stretch it

0.080

m. (b) Derive an expression for the elastic potential energy stored at extension

x

by integrating the spring force, and evaluate it for this extension. (c) A

0.50

kg mass hangs in equilibrium from the spring; determine the extension (

g = 9.8

m/s squared).

Show worked answer →

A 4-point spring FRQ linking force and energy.

(a) Force (1 point): $F = kx = (250)(0.080) = 20$ N.
(b) Energy (2 points): the work done against the spring is $U = \int_0^x kx'\,dx' = \tfrac{1}{2}kx^2$ . At $x = 0.080$ m: $U = \tfrac{1}{2}(250)(0.080)^2 = 0.80$ J.
(c) Equilibrium extension (1 point): the spring force balances the weight, $kx = mg$ , so $x = \dfrac{mg}{k} = \dfrac{(0.50)(9.8)}{250} = 0.0196$ m, about $2.0$ cm.

Markers reward integrating $kx$ to get $\tfrac{1}{2}kx^2$ and balancing $kx = mg$ for the hanging mass.

AP 2021 (style)1 marksSection I (multiple choice). If the extension of an ideal spring is doubled, the elastic potential energy stored becomes... (A) unchanged (B) doubled (C) tripled (D) quadrupled. Justify your reasoning.

Show worked answer →

A 1-point conceptual MCQ. The answer is (D).

The elastic potential energy is $U = \tfrac{1}{2}kx^2$ , proportional to the square of the extension. Doubling $x$ multiplies $U$ by $2^2 = 4$ . (By contrast the spring force $F = kx$ would only double.) The trap is to apply the linear scaling of the force to the quadratic energy.

Related dot points

Sources & how we know this

AP Physics C: Mechanics Course and Exam Description — College Board (2024)