Elastic Potential Energy Calculator

GCSE PhysicsAQAEdexcelOCRFoundation & Higher

Find elastic potential energy, the spring constant or the extension using Ee = ½ k e². Enter the extension in cm or mm if that's what you measured; the solver converts it to metres.

Quick answer

Elastic potential energy = 0.5 × spring constant × extension². Use k in N/m and extension in metres to get joules (J). For example, a spring with k = 40 N/m stretched by 0.05 m stores Ee = 0.5 × 40 × 0.05² = 0.05 J.

By Mehmood Kakar · Updated

Ee = ½ × k × e²
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How to calculate elastic potential energy

Elastic potential energy is the energy stored in a stretched or compressed spring. On your equation sheet it is written as:

elastic potential energy = 0.5 × spring constant × (extension)²

  • Ee is elastic potential energy in joules (J)
  • k is the spring constant in newtons per metre (N/m)
  • e is the extension in metres (m)

The extension is how much longer the spring gets, not its total length. A 10 cm spring stretched to 14 cm has an extension of 4 cm, which is 0.04 m.

The equation only works up to the limit of proportionality. Stretch a spring past that point and it no longer obeys F = ke, so Ee = ½ke² stops giving the right answer.

The work done stretching a spring equals the elastic potential energy it stores, as long as it is not stretched past its limit. A stiffer spring has a bigger spring constant, so it stores more energy for the same extension.

Because the extension is squared, doubling it stores four times the energy. Finding the extension needs a square root at the end; our rearranging guide shows the steps.

Common mistakes that cost marks

  1. Using the total length instead of the extension. Extension = stretched length − original length.
  2. Leaving the extension in cm. 5 cm is 0.05 m. Using 5 makes the answer 10,000 times too big, because the extension is squared.
  3. Forgetting to square the extension. Only e is squared, not the ½ or the k.
  4. Missing the square root when finding extension. e = √(2Ee ÷ k).
  5. Using the equation past the limit of proportionality. Beyond that point the spring no longer obeys Hooke’s law.

Practice questions

Try each one on paper first, then reveal the worked answer.

Q1A spring has a spring constant of 40 N/m. It is stretched by 0.050 m. Calculate the elastic potential energy stored.[2 marks]
  1. Ee = ½ × k × e²
  2. Ee = 0.5 × 40 × 0.050²
  3. 0.050² = 0.0025, so Ee = 0.5 × 40 × 0.0025
  4. Ee = 0.050 J
Q2A spring stores 0.48 J of elastic potential energy when it is stretched by 8.0 cm. Calculate its spring constant.[3 marks]
  1. Convert: e = 8.0 cm ÷ 100 = 0.080 m
  2. Rearrange: k = 2 × Ee ÷ e²
  3. k = (2 × 0.48) ÷ 0.080² = 0.96 ÷ 0.0064
  4. k = 150 N/m
Q3A spring with a spring constant of 200 N/m stores 0.25 J. Calculate its extension in cm.[3 marks]
  1. Rearrange: e = √(2 × Ee ÷ k)
  2. e = √(2 × 0.25 ÷ 200) = √0.0025
  3. e = 0.05 m
  4. e = 5.0 cm

Frequently asked questions

What is the elastic potential energy equation?

Elastic potential energy = 0.5 × spring constant × (extension)², written Ee = ½ke². Energy is in joules (J), spring constant in newtons per metre (N/m) and extension in metres (m).

When does Ee = ½ke² not work?

It only applies while the spring is within its limit of proportionality. If the spring is stretched further, force is no longer proportional to extension and the equation gives the wrong value.

How is elastic potential energy related to work done?

The work done to stretch a spring equals the elastic potential energy stored in it, provided the spring is not stretched past its limit of proportionality.

How do I find the extension?

Subtract the original length from the stretched length, then convert to metres. To find extension from energy, use e = √(2Ee ÷ k).