A spring 0.20 m long stretches to 0.25 m when a 10 N force is applied. Calculate the spring constant.
- Extension = 0.25 − 0.20 = 0.05 m.
- k = F ÷ e = 10 ÷ 0.05 = 200 N/m.
Answer 200 N/m
GCSE Combined Science revision · Physics: forces, waves and magnetism
Stretching, bending or compressing an object needs more than one force. Elastic deformation means the object returns to its original shape when the forces are removed; inelastic deformation means it does not. For a spring, the extension is directly proportional to the force applied, F = ke, up to the limit of proportionality. The spring constant k is in N/m. If the spring is not inelastically deformed, the work done stretching it is stored as elastic potential energy.
Watch out Extension is the increase in length, not the total length.
A spring 0.20 m long stretches to 0.25 m when a 10 N force is applied. Calculate the spring constant.
Answer 200 N/m
A spring with k = 50 N/m is stretched by 0.08 m. What force is applied, in newtons?
What is elastic deformation?
Beyond the limit of proportionality, extension and force are:
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