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Learning objectives
Core idea
Key formulae
Worked examples
Explore the graph
Common mistakes
Exam tip
Guided practice
Practice questions
Challenge
Summary
Quick quiz
Same topic, different board
Past paper questions

Learn9FM0Option · Further Mechanics 1Elastic strings and springsHooke's law and elastic energy

Elastic strings and springs · Subtopic 1 of 1

Elastic Strings and Springs

Tension in an elastic string or spring is $T=\frac{\lambda x}{l}$ , where $\lambda$ is the modulus of elasticity, $l$ the natural length and $x$ the extension. Stretching it stores elastic potential energy $\frac{\lambda x^2}{2l}$ , which joins kinetic and gravitational energy in the work–energy principle.

Cambridge 9231 Paper 3Edexcel YMA01 WME03Edexcel 9FM0 Option≈ 45 minRevision 1 · updated 2026-10-27

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Section 1 of 14

Learning objectives

Use Hooke's law $T=\frac{\lambda x}l$ .
Calculate elastic potential energy.
Solve equilibrium and energy problems with strings and springs.

Pause and think

A bungee cord stretches more the further the jumper falls. Why can't you use $v^2=u^2+2as$ for the whole fall?

Section 2 of 14

Core idea

Start hereThe bungee jump

A bungee jumper falls freely until the cord goes taut, then the cord pulls harder and harder until she stops. The force keeps changing, so the constant-acceleration equations fail. Energy saves the day: the cord simply stores energy as it stretches.

The big idea

Tension is proportional to extension: $T=\frac{\lambda x}l$ . The work done stretching the string is stored as elastic potential energy $\frac{\lambda x^2}{2l}$ , which you include alongside KE and GPE in energy equations.

1. Hooke's law

T=\frac{\lambda x}{l}

$\lambda$: modulus of elasticity (N): the tension when the string is doubled in length.
$l$: natural length.
$x$: extension (or compression for a spring).

Worked exampleHanging in equilibrium

A $2$ kg particle hangs on a string with $l=1.5$ m, $\lambda=49$ N. Find the extension ( $g=9.8$ ).

Goal: $x$ .
Strategy: Tension balances weight.

$T=mg=19.6$
Equilibrium.
$19.6=\frac{49x}{1.5}$
Hooke's law.
$x=\frac{19.6\times1.5}{49}=0.6$
Rearrange.

Answer

0.6

✓ Check

\frac{49\times0.6}{1.5}=19.6

✓.

2. Elastic potential energy

Work done stretching from

0

x

$W=\int_0^xT\,ds$
Work = force × distance, added over small steps since $T$ varies.
$=\int_0^x\frac{\lambda s}l\,ds$
Hooke's law at extension $s$ .
$=\frac{\lambda x^2}{2l}$
Integrate: the area of the triangle under the $T$ – $x$ graph.

EPE

=\frac{\lambda x^2}{2l}

Worked exampleReleased from a stretch

A $0.5$ kg particle hangs on a string ( $l=1$ m, $\lambda=49$ N) from $O$ . It is pulled to $1.5$ m below $O$ and released. Find its speed when the string goes slack.

Goal: $v$ at $1$ m below $O$ .
Strategy: Energy from the release point to the slack point.

$\text{EPE lost}=\frac{49(0.5)^2}{2(1)}=6.125$
Initial extension $0.5$ ; final $0$ .
$\text{GPE gained}=0.5(9.8)(0.5)=2.45$
It rises $0.5$ m.
$\tfrac12(0.5)v^2=6.125-2.45=3.675$
The rest becomes KE.
$v^2=14.7$
Solve.

Answer

v\approx3.83

m s

^{-1}

✓ CheckEnergy at the release point

6.125

J; at the slack point

2.45+3.675=6.125

J ✓.

Energy bookkeeping

Drawing diagram…

Energy changes form, but the total stays fixed without friction.

Common misconceptionExtension, not length Always subtract the natural length first. A

2

m string stretched to

2.5

m has

x=0.5

Edge caseSlack strings A string can't push. Once its length is below

l

it stores no energy and exerts no force; a spring keeps obeying Hooke's law in compression.

Pause and think

Two identical strings are joined end to end. What happens to the modulus and natural length?

Summary

$T=\frac{\lambda x}l$ .
EPE $=\frac{\lambda x^2}{2l}$ .
Use energy when tension varies.

Modulus of elasticity: $\lambda$ ; stiffness constant independent of length.
Natural length: Length with no tension.
Elastic potential energy: Energy stored by stretching or compressing.

HistoryRobert Hooke. Hooke published his law in 1678 as an anagram, "ceiiinosssttuv", which unscrambles to the Latin "ut tensio, sic vis": as the extension, so the force.

Fun factBungee maths. Bungee operators weigh every jumper and choose a cord so the lowest point stays safely above the ground: a direct use of elastic energy.

Section 3 of 14

Key formulae

Hooke's law

T=\frac{\lambda x}{l}

Elastic potential energy

E=\frac{\lambda x^2}{2l}

Work–energy

\text{KE}+\text{GPE}+\text{EPE}\ \text{conserved (no friction)}

TrickStrings vs springs. A string goes slack when its length drops below

l

(no thrust); a spring can be compressed and then pushes back with the same law.

Section 4 of 14

Worked examples

Example 1 · Equilibrium0 / 3 steps

A particle of mass

2

kg hangs on an elastic string of natural length

1.5

m and modulus

49

N. Find the extension.

Example 2 · Energy0 / 2 steps

A spring of natural length

0.5

m and modulus

40

N is compressed by

0.1

m against a

0.2

kg ball on a smooth table and released. Find the ball's speed.

Section 5 of 14

Explore the graph

Elastic energy is the area under T against x

Loading the interactive graph…

Try this

Here $k$ is $\lambda$ , $c$ the natural length and $s$ the extension. The shaded triangle has area $\frac12\cdot s\cdot\frac{\lambda s}l=\frac{\lambda s^2}{2l}$ .
Double $s$ : the energy quadruples.

Work done stretching = area under the tension graph.

In the real worldCar suspension. Car springs store energy as the wheel hits a bump and release it slowly through dampers, so the passengers barely notice.

Section 6 of 14

Common mistakes

IncorrectUsing the stretched length as

x

Correct

x=

length

-\ l

Hooke's law uses the extension.

IncorrectEPE

=\frac{\lambda x}{2l}

Correct

\frac{\lambda x^2}{2l}

Energy is the area under

T

, so

x

is squared.

IncorrectUsing

suvat

while the string is taut

CorrectUse energy

Acceleration varies.

Me stretching the string twice as far

Elastic energy: four times, actually.

Section 7 of 14

Exam tip

Exam tipChoose a zero level. Pick the GPE zero at the lowest point of the motion; it keeps signs simple.

Section 8 of 14

Guided practice

SA-FM-EL-G012 marksL1 FoundationCalculatorHooke's law

An elastic string has natural length

2

m and modulus

30

N. Find the tension when it is stretched to

2.5

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-FM-EL-0012 marksL2 StandardCalculatorHooke's law

2

kg particle hangs in equilibrium on an elastic string of natural length

1.5

m and modulus

49

N. Find the extension in metres (

g=9.8

Salam Academy Original Practice · Proposed mark scheme

SA-FM-EL-0022 marksL1 FoundationCalculatorElastic energy

Find the elastic potential energy in a spring of natural length

0.5

m and modulus

40

N compressed by

0.1

Salam Academy Original Practice · Proposed mark scheme

SA-FM-EL-0032 marksL2 StandardCalculatorEnergy problems

The spring in the last question launches a

0.2

kg ball along a smooth table. Find its speed in m s

^{-1}

Salam Academy Original Practice · Proposed mark scheme

SA-FM-EL-0042 marksL2 StandardCalculatorHooke's law

A string of natural length

0.8

m stretches to

1

m when a

0.5

kg mass hangs from it. Find the modulus

\lambda

(

g=9.8

Salam Academy Original Practice · Proposed mark scheme

More questions on every skill in this subtopic are in the question bank under Practise.

Section 10 of 14

Challenge

SA-FM-EL-0054 marksL4 ChallengingCalculatorEnergy problems

0.5

kg particle hangs on an elastic string (natural length

1

m, modulus

49

N) fixed at

O

. It is pulled down to

1.5

m below

O

and released. Find its speed in m s

^{-1}

when the string becomes slack (

g=9.8

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

$T=\frac{\lambda x}l$ .
EPE $=\frac{\lambda x^2}{2l}$ .
Energy conservation for variable tension.

Section 12 of 14

Quick quiz

Q1 · SA-FM-EL-Q011 markL1 FoundationNon-calculatorElastic energy

If the extension of an elastic string is tripled, the stored energy

Salam Academy Original Practice · Proposed mark scheme

Section 13 of 14

Same topic, different board

Specification: A level Further Mathematics 9FM0
Where: Further Mechanics 1 · Elastic strings and springs
Expected: Hooke's law, elastic potential energy, the work–energy principle with elastic strings and springs.
Calculator: Permitted; show enough working to earn the method marks.
Typical style: Edexcel often opens with a routine part and builds to a "show that" or "prove" part.
Open in: YMA01 · WME03 9FM0 · Option

Section 14 of 14

Past paper questions

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Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Pearson Edexcel 9FM0, First teaching 2017 (official specification).