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Solid Deformation · 8.1

Hooke's Law & Extension

Separate compression from tension, measure extension from the natural length, and use Hooke's law only inside its linear range.

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01

Build the model

Connect the load, geometry, and material response.

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.

Simple definition
Hooke's law says a spring's force is directly proportional to its extension while the spring remains in its proportional range.
Example
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.
Extensionx = L − L₀

How much longer the object got: stretched length minus natural length (negative when squashed).

Metres; x may be negative in compression

Hooke's lawF = kx

Within limits, a spring's pull is proportional to its stretch — double the stretch, double the force; k measures its stiffness.

Valid only in the proportional region

Graph gradientk = ΔF/Δx

The slope of the force–extension graph is the stiffness: a steeper line means a stiffer spring.

On a force–extension graph

F
tensile or compressive forcenewton, N
x
extension or compressionmetre, m
k
spring constantN m⁻¹
01

Measure the change

Do not substitute the final length for x. Subtract the natural length first.

02

Read the gradient

A steeper F–x line means a larger k and therefore a stiffer spring.

03

Know the boundary

Beyond the proportional limit, a curved graph means one constant k no longer describes the sample.

02

Change one input at a time

See the equation become a responsive graph.

Hooke model validStraight-line region
Force against extension with a user-set proportional limitextension x (m) on the horizontal axis and force F (N) on the vertical axis. Use the controls to change the highlighted point.000.0340.0680.08120.11160.1420extension x (m)force F (N)
Current force
6 N
Secant F/x
120 N m⁻¹
Model decision
Use F = kx

Past the selected limit, the curve shows only that one constant k is no longer sufficient; it is not a universal material law.

03

Catch the common trap

Predict before calculating.

Two springs are shown on the same force–extension axes. Which spring is stiffer?

Choose an answer, then compare the explanation with your model.

04

Worked examples

Convert units, choose the model, then check its range.

EasyA spring with k = 25 N/m is stretched 0.12 m. What force does it exert?
  1. F = kx = 25 × 0.12.
  2. F = 3.0 N pulling back toward natural length.

AnswerF = 3.0 N

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.
  1. x = 23.0 cm − 18.0 cm = 5.0 cm.
  2. Convert before calculating: x = 0.050 m.
  3. Use k = F/x = 6.0/0.050.

Answerx = 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.
  1. Extra 5.0 N produced extra 4.0 cm: k = 5.0 ÷ 0.040 = 125 N/m.
  2. At 4.0 N the extension is x = F/k = 4.0 ÷ 125 = 3.2 cm.
  3. Natural length = 30.0 − 3.2 = 26.8 cm.

Answerk = 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.
  1. Parallel pair: 2k = 600 N/m. In series with k: 1/k(total) = 1/600 + 1/300 → k(total) = 200 N/m.
  2. Load F = 6.0 × 9.81 ≈ 58.9 N.
  3. x = F/k = 58.9 ÷ 200 ≈ 0.29 m.

Answerx ≈ 0.29 m

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.

  1. 01

    InspectRead the figure comment.

  2. 02

    TraceFollow labels, arrows and axes.

  3. 03

    AnswerWork one part at a time.

  4. 04

    CheckReveal hints and marking points.

A Level

01Fig. 8.1Work, energy and power · Deformation of solidsA Level
A loaded steel wire hanging from a fixed supportfixed supportoriginal length2.50 msteel wirediameter 0.56 mm45 Nnot to scale

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.

5 Work, energy and power6 Deformation of solidsOpen A Level question · AS Level foundations Q8
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.

  1. aCalculate Calculate the cross-sectional area of the wire labelled in Fig. 8.1.

    recall2 marks

    Check answer 2 marks
    1. A = πd²/4 with d = 0.56 × 10⁻³ m
    2. A = 2.5 × 10⁻⁷ m²

  2. 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.

    routine3 marks

    Check answer 3 marks
    1. maximum load = stress × area = 8.0 × 10⁸ × 2.46 × 10⁻⁷
    2. = 2.0 × 10² N
    3. factor = 197 / 45 = 4.4

  3. 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.

    demanding3 marks

    Check answer 3 marks
    1. doubling the diameter makes the cross-sectional area 4 times greater
    2. for the same load the stress, and hence the strain, is one quarter of its previous value
    3. the extension becomes one quarter of its previous value

  4. 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.

    top of the paper4 marks

    Check answer 4 marks
    1. volume of wire = 2.46 × 10⁻⁷ × 2.50 = 6.2 × 10⁻⁷ m³
    2. weight of wire = 7800 × 6.2 × 10⁻⁷ × 9.81 = 0.047 N
    3. this is about 0.1% of 45 N, and only the part of the wire below a given point adds to the tension there
    4. 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
  1. stress = 750 / 1.1 × 10⁻⁴ = 6.8 × 10⁶ Pa
  2. strain = 0.16 / 12 = 0.013
  3. E = stress / strain = 5.1 × 10⁸ Pa