Elastic strain energy

Area under the graph

The work done in stretching = area under the force-extension graph. If the deformation is elastic, this energy is stored as elastic strain energy and can be recovered.

Linear graphs

When Hooke's law applies, the graph is a straight line through the origin and the area is a triangle:

E_el = (1/2)FΔx

Using F = kΔx this can also be written E_el = (1/2)k(Δx)².

Non-linear graphs

If the graph curves, find the area by counting squares (or splitting into strips and trapezia). Work out the energy one square represents (its force value x its extension value) and multiply by the number of squares. If the material deforms plastically, not all the work done is recovered: some is dissipated as thermal energy.

Worked example

A spring of stiffness 400 N/m is stretched by 0.050 m. Find the elastic strain energy stored.

1. F = kΔx = 400 x 0.050 = 20 N
2. E = (1/2)FΔx = 0.5 x 20 x 0.050

Answer: E = 0.50 J

Check yourself

1. A force of 12 N stretches a wire by 0.030 m (within the limit of proportionality). Find the energy stored, in J.
   Show answer

   0.18 J. E = (1/2)FΔx = 0.5 x 12 x 0.030 = 0.18 J.

2. A spring of stiffness 250 N/m is stretched by 0.040 m. Find the energy stored, in J.
   Show answer

   0.2 J. E = (1/2)k(Δx)² = 0.5 x 250 x 0.0016 = 0.20 J.

3. What does the area under a force-extension graph represent?
   • The stiffness
   • The Young modulus
   • The breaking force
   • The work done in stretching (elastic strain energy if elastic)
   Show answer

   The work done in stretching (elastic strain energy if elastic). Area = force x extension = work done.

4. The area under a curved force-extension graph covers 36 squares. Each square is 0.50 N high and 1.0 mm wide. Find the work done in J.
   Show answer

   0.018 J. One square = 0.50 x 1.0 x 10^-3 = 5.0 x 10^-4 J; 36 squares = 0.018 J.

5. A catapult stores 2.0 J of elastic strain energy and launches a 0.020 kg stone. If all the energy becomes kinetic energy, find the stone's speed in m/s, to 3 s.f.
   Show answer

   14.1 m/s. (1/2)mv² = 2.0, so v = sqrt(4.0 / 0.020) = sqrt(200) = 14.1 m/s.

6. The extension of a spring obeying Hooke's law is doubled. What happens to the energy stored?
   • It becomes 4 times as large
   • It doubles
   • It halves
   • It stays the same
   Show answer

   It becomes 4 times as large. E = (1/2)k(Δx)², so energy is proportional to the square of the extension.

7. A spring stores 0.90 J when stretched by 0.060 m. Find its stiffness in N/m.
   Show answer

   500 N/m. k = 2E / (Δx)² = 1.8 / 0.0036 = 500 N/m.

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Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics specification (XPH11/YPH11), Issue 3
• Pearson qualifications: International Advanced Levels Physics (2018)

Last updated 7 November 2026. Spotted a mistake? Tell us.