Chapter 14 · Solid Deformation · Lesson 8.1
Hooke’s Law & Extension
See how a force changes a spring’s length.
Separate compression from tension, measure extension from the natural length, and use Hooke's law only inside its linear range.
A simple model. A defined range.
Lesson Overview
Hooke's law says a spring's force is directly proportional to its extension while the spring remains in its proportional range.
If a spring needs 4 N to extend 2 cm, the same spring needs 8 N to extend 4 cm while Hooke's law still applies.
Learning Objectives
- Measure extension from natural length
- Convert lengths to SI units
- Use F = kx within its proportional range
- Determine stiffness from a graph
- Distinguish load and restoring force
- Separate proportional and elastic limits
Key Concepts
Interactive Spring Lab
Change one input. Predict the response before reading the force.
- Holding load
- 6 N
- Spring restoring force
- -6 N
- Elastic energy
- 0.15 J
- Secant Fload / x
- 120 N/m
x = 0.05 m. The holding load balances the restoring force at rest. The forces have equal magnitudes and opposite directions.
Force–extension graph
Blue: Fload = kx. Dashed boundary: 8 cm. Shaded region: response unknown. Axes stay fixed when k changes; convert centimetres to metres when calculating the gradient in N/m.
A quasi-static, lossless tensile model with an illustrative, adjustable limit—not measured spring data or a safety rating. The geometry shows the selected length only; the model does not predict permanent deformation.
Key Formulae
- Extension
- x = L − L₀Use the change from natural length, in metres. Negative x denotes compression.
- Holding load
- Fload = kxSigned external force for quasi-static loading. The lab uses tensile x ≥ 0.
- Restoring force
- Fspring = −kxThe spring force opposes displacement from its natural length; valid in the proportional range.
- Graph gradient
- k = ΔFload / ΔxForce vertical, extension horizontal. Stiffness has units N m⁻¹.
- Unit conversion
- 1 cm = 0.01 mConvert centimetres to metres before using k in N/m.
- Elastic energy
- U = ½kx²For a lossless linear spring, measured from its unloaded state. Explore more in lesson 8.2.
Length is not extension
A 20 cm spring stretched to 25 cm has x = 5 cm = 0.05 m, not 0.25 m.
Practice Questions
Measure the change. Then model the force.
A tensile force stretches a sample; a compressive force shortens it. Extension x is the signed change from the original length. For a spring in its proportional region, force and extension form a straight line through the origin, whose gradient is the spring constant k.
For tension, take x as positive. The external load holding the spring still is Fload = +kx; the spring’s restoring force on the pulling object is Fspring = −kx. For force magnitudes alone, write F = k|x|.
Proportionality has conditions
Use a constant k only over a verified straight-line range through the origin. The lab explores slow tensile loading, not oscillations, buckling or coil contact under compression.
A larger k means a stiffer spring, not necessarily a stronger one.
Measure the change
Do not substitute the final length for x. Subtract the natural length first.
Read the gradient
A steeper F–x line means a larger k and therefore a stiffer spring.
Know the boundary
Beyond the proportional limit, a curved graph means one constant k no longer describes the sample.
Two different limits.
Proportional limit
The end of the straight-line force–extension relationship. Beyond it, one constant k no longer describes the response.
Elastic limit
The boundary beyond which the sample does not fully recover its original dimensions when unloaded. You need unloading evidence to identify permanent deformation.
The elastic limit and proportional limit are related but not identical ideas. A material can stop being proportional before it stops returning fully to its original shape.
Two springs are shown on the same force–extension axes. Which spring is stiffer?
Choose an answer to test the idea.
Worked Examples
Assume ideal, negligible-mass springs, static loads and extensions within each spring’s proportional range. State the force direction separately from its magnitude.
EasyA spring with k = 25 N/m is stretched 0.12 m. What force does it exert?
- F = kx = 25 × 0.12.
- F = 3.0 N pulling back toward natural length.
F = 3.0 N in magnitude; signed restoring force −3.0 N (opposite the extension).
MediumA spring lengthens from 18.0 cm to 23.0 cm under a 6.0 N load. Find its extension and spring constant while Hooke's law applies.
- x = 23.0 cm − 18.0 cm = 5.0 cm.
- Convert before calculating: x = 0.050 m.
- Use k = F/x = 6.0/0.050.
x = 0.050 m and k = 120 N m⁻¹
HardA spring is 30.0 cm long under a 4.0 N load and 34.0 cm under a 9.0 N load. Find k and the natural length.
- Extra 5.0 N produced extra 4.0 cm: k = 5.0 ÷ 0.040 = 125 N/m.
- At 4.0 N, x = F/k = 4.0 ÷ 125 = 0.032 m = 3.2 cm.
- Natural length = 30.0 − 3.2 = 26.8 cm.
k = 125 N/m; L₀ = 26.8 cm
ChallengingThree identical springs (k = 300 N/m) support a 6.0 kg mass: two in parallel, joined in series to the third. Find the total extension.
- Parallel pair: 2k = 600 N/m. In series with k: 1/k(total) = 1/600 + 1/300 → k(total) = 200 N/m.
- Load F = 6.0 × 9.81 ≈ 58.9 N.
- x = F/k = 58.9 ÷ 200 ≈ 0.29 m.
x ≈ 0.29 m
Test a spring.
- Secure the spring and measure its unloaded length L₀. Use a pointer and read at eye level.
- Add known loads in small steps. Once still, calculate F = mg and record L, then x = L − L₀.
- Plot force vertically against extension horizontally. Use a best-fit line in the straight region to estimate k.
- Unload gradually and compare lengths to test recovery. Repeat readings and discuss uncertainty.
Work safely
Use a stable stand, secure masses, a tray beneath the load and eye protection. Stay within the apparatus’s rated load; do not deliberately load it to failure.
The limit slider is a teaching assumption. It is not a test result or a safe working load for a real spring.
Springs and structures in the real world.
Spring balances
Within a calibrated linear range, extension can indicate force. Zero the scale before adding a load.
Suspension systems
Springs deform under changing loads. Dampers dissipate energy; an ideal spring alone does not model the full ride.
Cables and structures
Small elastic extensions depend on material and geometry. Stiffness alone cannot tell you when a structure will fail.
Explore Young modulus →Exam diagrams for this topic1 figure to inspect and practiseQuestions, hints and marking points in one compact subsection.
See it. Read it. Work it.
These figures come from GioPhysics practice papers. Open one, decode the drawing, work the guided questions, then follow its link to the full paper question.
- 01
InspectRead the figure comment.
- 02
TraceFollow labels, arrows and axes.
- 03
AnswerWork one part at a time.
- 04
CheckReveal hints and marking points.
A Level
01Fig. 8.1Work, energy and power · Deformation of solidsA Level
Figure comment
Fig. 8.1A steel wire hangs vertically from a rigid support drawn as a hatched ceiling, and a block is attached to its lower end. An arrow starting at the centre of that block points vertically downwards and is labelled 45 N. A dimension line to the left of the wire marks its original length as 2.50 m, and a leader line to the wire labels it as a steel wire of diameter 0.56 mm. The figure is not to scale.
Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.
Guided questions 5 parts
Reading cue. The 45 N is the weight of the block, so it equals the tension only if the wire's own weight is ignored; 0.56 mm is a diameter, and 2.50 m is the length before loading.
aCalculate Calculate the cross-sectional area of the wire labelled in Fig. 8.1.
Check answer 2 marks
- A = πd²/4 with d = 0.56 × 10⁻³ m
- A = 2.5 × 10⁻⁷ m²
bDetermine Steel of this type breaks at a tensile stress of 8.0 × 10⁸ Pa. Determine the greatest load this wire could support, and the factor by which the 45 N load could be increased before the wire breaks.
Check answer 3 marks
- maximum load = stress × area = 8.0 × 10⁸ × 2.46 × 10⁻⁷
- = 2.0 × 10² N
- factor = 197 / 45 = 4.4
cDeduce The wire is replaced by one of the same steel and the same original length but of diameter 1.12 mm, and the same 45 N load is hung from it. Deduce the factor by which the extension changes.
Check answer 3 marks
- doubling the diameter makes the cross-sectional area 4 times greater
- for the same load the stress, and hence the strain, is one quarter of its previous value
- the extension becomes one quarter of its previous value
dSuggest The density of steel is 7800 kg m⁻³. Suggest, with a supporting calculation, whether taking the tension in the wire to be 45 N along its whole length is justified.
Check answer 4 marks
- volume of wire = 2.46 × 10⁻⁷ × 2.50 = 6.2 × 10⁻⁷ m³
- weight of wire = 7800 × 6.2 × 10⁻⁷ × 9.81 = 0.047 N
- this is about 0.1% of 45 N, and only the part of the wire below a given point adds to the tension there
- so treating the tension as 45 N throughout introduces no significant error
Transfer challenge
A climbing rope of unstretched length 12 m and cross-sectional area 1.1 × 10⁻⁴ m² stretches by 0.16 m when a climber of weight 750 N hangs at rest from it. Calculate the Young modulus of the rope material.
Check answer 3 marks
- stress = 750 / 1.1 × 10⁻⁴ = 6.8 × 10⁶ Pa
- strain = 0.16 / 12 = 0.013
- E = stress / strain = 5.1 × 10⁸ Pa
Three checks before you move on.
- Subtract natural length and convert x to metres.
- Name the force: holding load or spring restoring force.
- Check the proportional range; do not confuse it with elastic recovery.
Quick Quiz
Five questions. Answers stay in this open page only.
Downloads & Resources
Try a Game
Bridge Load TesterExplore how cable diameter changes stress and extension under load.
Play now →