A wire of cross-sectional area 2.5 × 10^-7\ m^2 carries a load of 50\ N. Find the stress. [2 marks]

WJEC Eduqas Eduqas 2018-style practice: A vertical steel wire of length 2.5\ m and diameter 0.80\ mm supports a load of 45\ N and stretches by 1.4\ mm. Calculate the stress, the strain and the Young modulus of the steel.

Cross-sectional area: A = π r^2 = π(0.40 × 10^-3)^2 = 5.03 × 10^-7\ m^2. Stress: σ = FA = 455.03 × 10^-7 = 8.95 × 10^7\ Pa. Strain: = LL = 1.4 × 10^-32.5 = 5.6 × 10^-4 (no units). Young modulus: E = σ = 8.95 × 10^75.6 × 10^-4 = 1.6 × 10^11\ Pa. Markers reward the area from the radius, stress about 9.0 × 10^7\ Pa, strain about 5.6 × 10^-4, and the Young modulus about 1.6 × 10^11\ Pa.

EnglandPhysicsSyllabus dot point

How do solids deform under load, and what do stress, strain and the Young modulus tell us about a material?

Solids under stress: Hooke's law and the force constant, stress, strain and the Young modulus, elastic strain energy, and the contrast between ductile, brittle and polymeric behaviour.

A focused answer to the Eduqas A-Level Physics Component 2 solids under stress content, covering Hooke's law and the force constant, the definitions of stress, strain and the Young modulus, elastic strain energy as an area, and the contrast between ductile, brittle and polymeric materials.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Hooke's law states that the extension of a spring is proportional to the force, $F = k\Delta x$ , where $k$ is the force constant, up to the limit of proportionality. Stress is force per unit area, $\sigma = \frac{F}{A}$ (in pascals); strain is extension per unit original length, $\varepsilon = \frac{\Delta L}{L}$ (no units). The Young modulus is stress divided by strain in the linear region, $E = \frac{\sigma}{\varepsilon}$ , a measure of stiffness and a property of the material. Elastic strain energy is the area under a force-extension graph, $E = \frac{1}{2}F\Delta x = \frac{1}{2}k(\Delta x)^2$ . Ductile materials (such as copper) deform plastically before breaking, brittle materials (such as glass) snap with little plastic deformation, and polymeric materials (such as rubber) show large, non-linear, often hysteretic extensions.

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What this dot point is asking

Eduqas wants you to state Hooke's law and define the force constant, define stress, strain and the Young modulus, find the elastic strain energy as the area under a force-extension graph, and contrast the stress-strain behaviour of ductile, brittle and polymeric materials.

The answer

Hooke's law and the force constant

Stress, strain and the Young modulus

Elastic strain energy

Ductile, brittle and polymeric behaviour

The Young modulus from a wire experiment

A wire of original length $1.80\ \text{m}$ and cross-sectional area $2.0 \times 10^{-7}\ \text{m}^2$ extends by $0.90\ \text{mm}$ under a load of $40\ \text{N}$ . Find the Young modulus.

step 1: Calculate the stress

$\sigma = \dfrac{F}{A} = \dfrac{40}{2.0 \times 10^{-7}} = 2.0 \times 10^{8}\ \text{Pa}$ .

step 2: Calculate the strain

$\varepsilon = \dfrac{\Delta L}{L} = \dfrac{0.90 \times 10^{-3}}{1.80} = 5.0 \times 10^{-4}$ .

step 3: Divide to find the Young modulus

$E = \dfrac{\sigma}{\varepsilon} = \dfrac{2.0 \times 10^{8}}{5.0 \times 10^{-4}} = 4.0 \times 10^{11}\ \text{Pa}$ .

Examples in context

The Young modulus governs the choice of material in construction and engineering: steel beams, suspension cables and aircraft components are all selected for the right combination of stiffness, strength and density. Elastic strain energy is the basis of springs, trampolines, archery bows and the energy returned by running shoes. The hysteresis of rubber explains why car tyres heat up and why rubber is used for vibration damping and shock absorption.

Try this

Q1. State Hooke's law. [1 mark]

Cue. Up to the limit of proportionality, the extension is directly proportional to the applied force.

Q2. A wire of cross-sectional area $2.5 \times 10^{-7}\ \text{m}^2$ carries a load of $50\ \text{N}$ . Find the stress. [2 marks]

Cue. $\sigma = \frac{F}{A} = \frac{50}{2.5 \times 10^{-7}} = 2.0 \times 10^{8}\ \text{Pa}$ .

Q3. State the difference between a ductile and a brittle material. [2 marks]

Cue. A ductile material deforms plastically over a large extension before breaking; a brittle material snaps with little or no plastic deformation.

Exam-style practice questions

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

Eduqas 20185 marksA vertical steel wire of length

2.5\ \text{m}

and diameter

0.80\ \text{mm}

supports a load of

45\ \text{N}

and stretches by

1.4\ \text{mm}

. Calculate the stress, the strain and the Young modulus of the steel.

Show worked answer →

Cross-sectional area: $A = \pi r^2 = \pi(0.40 \times 10^{-3})^2 = 5.03 \times 10^{-7}\ \text{m}^2$ .

Stress: $\sigma = \dfrac{F}{A} = \dfrac{45}{5.03 \times 10^{-7}} = 8.95 \times 10^{7}\ \text{Pa}$ .

Strain: $\varepsilon = \dfrac{\Delta L}{L} = \dfrac{1.4 \times 10^{-3}}{2.5} = 5.6 \times 10^{-4}$ (no units).

Young modulus: $E = \dfrac{\sigma}{\varepsilon} = \dfrac{8.95 \times 10^{7}}{5.6 \times 10^{-4}} = 1.6 \times 10^{11}\ \text{Pa}$ .

Markers reward the area from the radius, stress about $9.0 \times 10^{7}\ \text{Pa}$ , strain about $5.6 \times 10^{-4}$ , and the Young modulus about $1.6 \times 10^{11}\ \text{Pa}$ .

Eduqas 20214 marksA spring obeying Hooke's law has a force constant of

25\ \text{N m}^{-1}

and is stretched by

0.080\ \text{m}

. Calculate the force needed and the elastic strain energy stored, and explain what happens to this energy when the spring is released.

Show worked answer →

Force from Hooke's law: $F = k\Delta x = 25 \times 0.080 = 2.0\ \text{N}$ .

Strain energy is the area under the force-extension line: $E = \tfrac{1}{2}F\Delta x = \tfrac{1}{2}k(\Delta x)^2 = \tfrac{1}{2}(25)(0.080)^2 = \tfrac{1}{2}(25)(0.0064) = 0.080\ \text{J}$ .

When released, this stored elastic potential energy is transferred to kinetic energy of the spring and any attached mass (and eventually to heat as the oscillation is damped). Markers reward $F = 2.0\ \text{N}$ , the strain energy $0.080\ \text{J}$ , and identifying the transfer to kinetic energy.

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Sources & how we know this

Eduqas GCE AS/A Level Physics specification (A720QS) — WJEC Eduqas (2015)