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Forces and motion · Subtopic 4 of 6

Springs and Hooke's Law

A spring stretches in proportion to the load on it, F=kxF=kx, up to the limit of proportionality. The spring constant kk tells you how stiff it is. Extension is measured from the original length, and a load–extension graph is a straight line through the origin until the limit is passed.

Cambridge 5054 Paper 1Cambridge 5054 Paper 2Cambridge 0625 Papers 2, 4Cambridge 0625 Papers 1, 3Edexcel 4PH1 Paper 1PEdexcel 4PH1 Paper 2P≈ 25 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Measure extension correctly from a spring's natural length.
  • Use k=Fxk=\dfrac Fx to find the spring constant, force or extension.
  • Interpret load–extension graphs and identify the limit of proportionality.
  • Describe an experiment to investigate how a spring stretches.
Pause and think

One 11 N weight stretches a spring by 22 cm. What would you predict for 33 N? Would the pattern go on forever?

Section 2 of 14

Core idea

Start hereA spring balance

Hang 1 N on a spring and it stretches 4 cm; hang 2 N and it stretches 8 cm. This simple doubling is why a spring can be marked as a scale. It works until you pull too hard.

The big idea

Hooke's law: F=kxF=kx up to the limit of proportionality. Extension = stretched length − original length.

1. Extension

Load–extension graph for a spring
2468101214123456780extension (cm)load (N)limit of proportionalitystraight: F = kxcurves: stretches easily
In the straight part, doubling the load doubles the extension. Here k=4.88=0.6k=\dfrac{4.8}{8}=0.6 N/cm.
Extension = new length − original length
extension xloadno load
Always measure from the unloaded length, not from zero on the ruler.
F=kxF=kx
where
FF
load (N).
kk
spring constant (N/m): stiffness.
xx
extension (m).
Worked exampleFinding k

Original length 12.0 cm; with 4 N it is 28.0 cm.

Goal
kk; load for a 10 cm extension.
Strategy
Extension, then k=F/xk=F/x.
  1. x=28.0−12.0=16.0 cm=0.16 mx=28.0-12.0=16.0\text{ cm}=0.16\text{ m}
    Extension, in metres.
  2. k=40.16=25 N/mk=\frac{4}{0.16}=25\text{ N/m}
    Rearrange.
  3. F=25×0.10=2.5 NF=25\times0.10=2.5\text{ N}
    Use.
Answer2525 N/m; 2.52.5 N.
✓ Check10 cm is 58\frac58 of 16 cm, and 2.52.5 N is 58\frac58 of 4 N ✓.
Common misconceptionLength instead of extension Use 28−12=1628-12=16 cm, not 2828 cm.
Edge caseBeyond the limit Past the limit of proportionality the graph curves; past the elastic limit the spring stays permanently stretched.
Pause and think

Two identical springs side by side share a load. Effect on extension?

Summary

  • Extension = new − original.
  • F=kxF=kx, straight line through the origin.
Key words
Spring constant
Force per unit extension.
Limit of proportionality
End of the straight-line region.
HistoryHooke's anagram. In 16761676 Robert Hooke published his law as an anagram, "ceiiinosssttuv", to claim priority without revealing it. Two years later he gave the answer: ut tensio, sic vis, "as the extension, so the force".
Fun factSprings in your bed and your watch. A mechanical watch keeps time with a tiny hairspring vibrating back and forth. Because of Hooke's law, every swing takes the same time, whatever its size.
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