The idea
The restoring force acts toward the equilibrium position; in a signed coordinate system this is written F = −kx. Spring constant has units N/m. A steeper force–extension graph means a stiffer spring. The elastic potential energy stored in a spring obeying Hooke’s law is ½kx², equal to the area under its force–extension graph.
Worked example
A spring of natural length 0.20 m stretches to 0.23 m under a force of 6 N. Find its spring constant.
- Extension x = 0.23 − 0.20 = 0.03 m.
- k = F/x = 6 ÷ 0.03 = 200 N/m.
- Stored energy = ½ × 200 × 0.03² = 0.09 J.
A mistake to avoid
Use extension, not the total length of the stretched spring. Convert centimetres to metres before calculating.
Check your understanding
For the same Hookean spring, what happens to stored energy if extension doubles?
Show the explanation
It becomes four times as large because elastic potential energy is proportional to the square of extension.
Explore the lesson
This guide connects to Force Equations — Friction, Springs & Fluids in the IB Physics course. The first two published lessons per subject are free previews; further interactive lessons require an account.