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AQA · GCSE · Physics · Higher

Elastic potential energy stored in the spring calculation

Q02.2 4 marks
Ee=12ke2E_e = \frac{1}{2} k e^2

When the gymnast lands on the floor, one of the springs compresses by 1.2 cm.

text{spring constant} = 8500\ text{N/m}

Calculate the elastic potential energy stored in the spring. Give the unit.

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Marking points

  1. 1 Converts 1.2 cm to 0.012 m.
  2. 2 Uses Eₑ = ½ke² with the supplied spring constant and compression.
  3. 3 Obtains 0.612 (0.61 allowed).
  4. 4 Supplies J or joule in the separate Unit field.

Full-mark answer

Convert the compression: e = 1.2 cm = 0.012 m. Eₑ = ½ke² = 0.5 × 8500 × 0.012² = 0.612 J.

Why this answer loses marks

I used the compression in centimetres directly.

The deformation must be converted to metres before it is squared in the elastic-energy equation.

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