A wire of area 1.0 × 10^-6\ m^2 carries a load of 80\ N. Find the stress. [1 mark]

A spring of force constant 200\ N m^-1 is stretched by 0.050\ m. Find the strain energy stored. [2 marks]

OCR OCR 2018-style practice: A steel wire of length 2.5\ m and diameter 0.50\ mm stretches by 1.8\ mm under a load of 45\ N. Calculate the Young modulus of the steel.

Cross-sectional area: A = (π d^2)/(4) = π (0.50 × 10^-3)^24 = 1.96 × 10^-7\ m^2. Stress: σ = (F)/(A) = 451.96 × 10^-7 = 2.29 × 10^8\ Pa. Strain: = ( L)/(L) = 1.8 × 10^-32.5 = 7.2 × 10^-4. Young modulus: E = (σ)/() = 2.29 × 10^87.2 × 10^-4 = 3.2 × 10^11\ Pa. Markers reward the area, stress, strain and the final value about 3 × 10^11\ Pa (close to the accepted value for steel).

EnglandPhysicsSyllabus dot point

How do materials deform under load, and what do stress, strain and the Young modulus tell us?

Materials: Hooke's law and the force constant, elastic and plastic deformation, stress, strain and the Young modulus, elastic strain energy, and the behaviour of ductile, brittle and polymeric materials.

A focused answer to the OCR H556 forces and motion content on materials, covering Hooke's law and the force constant, elastic and plastic deformation, stress, strain and the Young modulus, elastic strain energy as the area under a force-extension graph, and ductile, brittle and polymeric behaviour.

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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What this dot point is asking

OCR wants you to apply Hooke's law and the force constant, distinguish elastic from plastic deformation, define and calculate stress, strain and the Young modulus, find elastic strain energy from a force-extension graph, and describe the behaviour of ductile, brittle and polymeric materials.

The answer

Hooke's law and the force constant

Elastic and plastic deformation

Stress, strain and the Young modulus

The Young modulus is the gradient of the straight-line part of a stress-strain graph. Steel has a high Young modulus (very stiff); rubber has a low one.

Elastic strain energy and material behaviour

Stretching a copper wire

A copper wire of natural length $1.6\ \text{m}$ and cross-sectional area $2.0 \times 10^{-7}\ \text{m}^2$ supports a $50\ \text{N}$ load. The Young modulus of copper is $1.3 \times 10^{11}\ \text{Pa}$ . Find the extension.

step 1: Stress

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

step 2: Strain

$\varepsilon = \frac{\sigma}{E} = \frac{2.5 \times 10^{8}}{1.3 \times 10^{11}} = 1.92 \times 10^{-3}$ .

step 3: Extension

$\Delta L = \varepsilon L = 1.92 \times 10^{-3} \times 1.6 = 3.1 \times 10^{-3}\ \text{m}$ (about $3.1\ \text{mm}$ ).

Examples in context

Engineers select materials by their Young modulus and breaking stress: steel for stiff load-bearing beams, carbon fibre for a high stiffness-to-weight ratio. Force-extension and stress-strain graphs are standard in the OCR practical work, where the Young modulus is found from the gradient and the area gives strain energy. Climbing ropes are deliberately designed to absorb energy by stretching, storing elastic strain energy and reducing the peak force on a falling climber.

Try this

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

Cue. Extension is directly proportional to the applied force, up to the limit of proportionality.

Q2. A wire of area $1.0 \times 10^{-6}\ \text{m}^2$ carries a load of $80\ \text{N}$ . Find the stress. [1 mark]

Cue. $\sigma = \frac{F}{A} = \frac{80}{1.0 \times 10^{-6}} = 8.0 \times 10^{7}\ \text{Pa}$ .

Q3. A spring of force constant $200\ \text{N m}^{-1}$ is stretched by $0.050\ \text{m}$ . Find the strain energy stored. [2 marks]

Cue. $E = \frac{1}{2}k(\Delta x)^2 = \frac{1}{2}(200)(0.050)^2 = 0.25\ \text{J}$ .

Exam-style practice questions

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

OCR 20185 marksA steel wire of length

2.5\ \text{m}

and diameter

0.50\ \text{mm}

stretches by

1.8\ \text{mm}

under a load of

45\ \text{N}

. Calculate the Young modulus of the steel.

Show worked answer →

Cross-sectional area: $A = \frac{\pi d^2}{4} = \frac{\pi (0.50 \times 10^{-3})^2}{4} = 1.96 \times 10^{-7}\ \text{m}^2$ .

Stress: $\sigma = \frac{F}{A} = \frac{45}{1.96 \times 10^{-7}} = 2.29 \times 10^{8}\ \text{Pa}$ .

Strain: $\varepsilon = \frac{\Delta L}{L} = \frac{1.8 \times 10^{-3}}{2.5} = 7.2 \times 10^{-4}$ .

Young modulus: $E = \frac{\sigma}{\varepsilon} = \frac{2.29 \times 10^{8}}{7.2 \times 10^{-4}} = 3.2 \times 10^{11}\ \text{Pa}$ .

Markers reward the area, stress, strain and the final value about $3 \times 10^{11}\ \text{Pa}$ (close to the accepted value for steel).

OCR 20213 marksA spring obeying Hooke's law has a force constant of

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

. It is stretched by

0.15\ \text{m}

. Calculate the elastic strain energy stored and state one assumption.

Show worked answer →

Elastic strain energy is the area under the force-extension graph, $E = \frac{1}{2}k(\Delta x)^2 = \frac{1}{2}(80)(0.15)^2 = \frac{1}{2}(80)(0.0225) = 0.90\ \text{J}$ .

Assumption: the spring obeys Hooke's law throughout (it is not stretched beyond its limit of proportionality), so the force-extension graph is a straight line.

Markers reward the formula, the value $0.90\ \text{J}$ , and a valid assumption.

Related dot points

Sources & how we know this

OCR AS and A Level Physics A (H156, H556) specification — OCR (2015)