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Hooke’s law and spring calculations

Use force and extension, find spring constants and elastic energy, and interpret force–extension data.

Practise

Elastic potential energy

A spring of constant 80 N m⁻¹ is stretched 5 cm from its unloaded length. The spring obeys Hooke’s law over this range. Find the elastic potential energy stored. Give your answer to 3 significant figures.

Treat the given values as exact unless an uncertainty is stated.

Use decimals or e notation, e.g. 4.2e-3. Enter the unit separately.

Hint

Use E = ½kx²; convert to metres before squaring.

Worked solution
  1. E = 0.5 × 80 × 0.05² J.
  2. Answer: 0.100 J (3 significant figures).
Another question
Question type: Elastic potential energy
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Foundation and Higher: this set covers content common to both tiers. Some calculations need several steps.

Force and extension

Extension is the change from unloaded length, measured in metres for F = kx. The force–extension gradient is k when force is vertical and extension horizontal. E = ½kx² applies over the Hooke’s-law range. The limit of proportionality and elastic limit are different: a curved relationship does not by itself prove permanent deformation.

Work through the practical method and sample results

GCSE scope

AQA 8463: 4.5.3; required practical 6 (selected); OCR Gateway J249: P2.3b–e; PM2.3i–iii; PAG P2 (selected). Tasks cover the selected skills listed above.

AQA specification · OCR Gateway specification