AS Level · Topic 6.1
A-Level 9702 / Topic 6 / AS

How much it stretches.

Pull on a wire and it lengthens. How much depends on the force, the dimensions and the material. Hooke law handles the spring; stress, strain and the Young modulus strip away the shape to describe the material itself.

The key idea

Below the limit of proportionality, extension is proportional to load: F = kx, with k the spring constant. To describe the material, use tensile stress σ = F / A and tensile strain ε = x / L. Their ratio is the Young modulus, E = stress / strain, a property of the material independent of the sample size.

limit of proportionality gradient = E stress σ strain ε
Fig. 1 — In the linear (Hooke) region the gradient of the stress–strain graph is the Young modulus, E = σ/ε
Section 01

Load it, watch it extend.

Add load to the wire and trace the force-extension graph. While it is straight the wire obeys Hooke law; the gradient is the spring constant. Read off the stress, strain and Young modulus as you go.

Section 02

From spring to material.

The spring constant describes one sample; the Young modulus describes the material.

RelationMeaningUnit
F = kxHooke lawspring constant k in N m⁻¹
σ = F / Atensile stresspascal (Pa)
ε = x / Ltensile strainno unit
E = σ / εYoung moduluspascal (Pa)
Stage 1 · Learn

Check what the sim just showed you

Four quick checks tied to this lesson. Each correct answer earns XP and lights this skill on your star map.

Quick check+10 XP

Hooke law states that, below the limit of proportionality, the extension of a spring is:

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A spring extends by 0.040 m under a load of 8.0 N. Its spring constant is:

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The tensile stress in a wire of cross-sectional area A carrying a tension F is:

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Tensile strain is defined as the:

Section 03

Measuring the Young modulus of a wire.

The standard experiment turns four measurements into one material constant.

Examiner trap

Keep the limits straight: the limit of proportionality is where the graph stops being straight, while the elastic limit is where permanent deformation begins; they are close but not the same. Stress uses the cross-sectional area (A = πd²/4 for a round wire), not the diameter or length. Strain is a ratio and has no units, so an answer with units of metres is wrong.

Stage 2 · Exam

Exam-style questions

Unlocks once the checks above are done. Worth more XP, written to AS Paper 1 and 2 standard.

Finish the checks above to unlock the exam questions
Exam style+20 XP

A wire of cross-sectional area 2.0 × 10⁻⁷ m² carries a tension of 40 N. The tensile stress is:

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A wire of original length 2.5 m stretches by 1.0 mm. The strain is:

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A material has a stress of 1.2 × 10⁸ Pa at a strain of 6.0 × 10⁻⁴. Its Young modulus is:

Skill unlocked

Stress, strain and the Young modulus, mastered.

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Go deeper · practice
Six original Cambridge-style questions
Spring constants from data, stress and strain calculations, the Young modulus from a graph, and the wire experiment. Attempt each, then reveal the worked solution.
Stage 3 · Paper 1 readiness
Deformation of solids · Paper 1 Practice
A bank of original multiple-choice questions across the whole topic, in the style of Paper 1. Start this once you are confident across Hooke law, stress and strain, and elastic energy.
Start Paper 1 Practice →