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Hooke's law and springs

Apply F = kx within the limit of proportionality, find k from a force–extension graph and combine springs in series and parallel.

  • GCE A-Level H2 Physics 2027

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Hang twice the load on a spring and it stretches twice as far, at least until it is stretched too much. That simple rule, Hooke’s law, lets you predict how far a spring stretches, find its stiffness from a graph and work out what happens when springs are joined. This lesson states Hooke’s law, its limits, and how to use it.

Hooke’s law

Within its limit of proportionality, the force needed to stretch a spring is proportional to its extension:

F = kx

k is the force constant of the spring, in N m⁻¹: the force needed per metre of extension. A stiffer spring has a larger k. The extension x is the increase in length from the spring’s natural, unloaded length: x = L-L₀. For a compressed spring, x is the decrease in length.

Extension is the change from natural lengthAn unloaded spring has natural length L0. The same spring is stretched to loaded length L. Extension x is L minus L0, shown between the original and new end positions. The applied force points right and the spring's restoring force on the pulling body points left.Measure from the same fixed endUnloadedLoadedL₀Lx = L − L₀Applied forceon springSpring force onpulling body
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Use x = L − L0, rather than the total loaded length L, in Hooke's law. The two force arrows belong to the spring and the pulling body respectively; they are not two forces on one body. Spring coils and lengths are schematic.

The spring pulls back on whatever stretches it with a force of the same size, always directed back towards its natural length.

For a load hanging at rest on a spring, the spring’s upward pull balances the weight: kx = mg.

Guided practice 1

Force constant from two lengths

About 3 min

Problem

A spring is 4.0 cm long when unloaded and 7.5 cm long when a force of 14 N stretches it, within its limit of proportionality. Find its force constant.

Find k

Unit: N m⁻¹

Hints

Hint 1: extension, then metres

The extension is the change in length, 7.5-4.0 cm. Convert it to metres before dividing.

Show solution step by step

  1. Extension

    Reason

    F = kx uses the extension, not the total length.

    Working

    x = 7.5-4.0 = 3.5 cm = 0.035 m

  2. Force constant

    Working

    k = F/x = 14/0.035 = 400 N m⁻¹

Check your understanding 1

A spring with force constant 250 N m⁻¹ is 0.180 m long unloaded and 0.212 m long under a load. A learner writes F = 250 × 0.212 = 53 N. What is wrong, and what is the force?

Show answer

F = kx uses the extension, not the total length. x = 0.212-0.180 = 0.032 m, so F = 250 × 0.032 = 8.0 N.

The limit of proportionality and the elastic limit

On a graph of force against extension, Hooke’s law gives a straight line through the origin, and its gradient is k. The line stays straight only up to the limit of proportionality; beyond it the graph curves and no single k applies.

The elastic limit is the greatest load from which the spring still returns to its natural length when the load is removed. Beyond it the spring is permanently stretched. The two limits are often close, but they are not the same thing: a spring can be past its limit of proportionality yet still return to its original length.

Force–extension graph with proportional and elastic limitsIn this schematic example, force is proportional to extension from the origin to point P. The material then remains elastic through a curved region until point E; beyond E, unloading may leave permanent extension.Extension, xForce, FPEF = kxlinear; gradient = knon-linear butstill elasticpermanent extensionmay remain
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P ends the straight Hooke’s-law region. E marks the limit of reversible deformation. This is a schematic example, not measured data; the curve and separation of P and E depend on the material.

Check your understanding 2

A spring’s force–extension graph starts to curve above 6 N, but the spring still returns to its original length after being loaded to 8 N. Which limit has it passed?

Show answer

The limit of proportionality, but not the elastic limit. Above 6 N the extension is no longer proportional to the force, but the spring was not permanently stretched.

Try it yourself 2

Read a force–extension graph

Minimal support

Problem

A spring’s force–extension graph is a straight line from the origin to the point x = 0.080 m, F = 16 N. Find the force constant, and the work done in stretching the spring to this extension.

Use the gradient and the area

Unit: N m⁻¹
Unit: J

Hints

Hint 1: gradient and area

For a straight line through the origin, k = F/x, and the area under it is the work done, (1/2)Fx.

Show solution step by step

  1. Gradient

    Reason

    The gradient of the straight line is the force constant.

    Working

    k = 16/0.080 = 200 N m⁻¹

  2. Area

    Working

    W = (1/2)(16)(0.080) = 0.64 J

Springs in series and in parallel

For light springs that obey Hooke’s law:

  • In parallel, both springs are stretched by the same amount, and they share the load: k_eff = k₁ + k₂. The pair is stiffer than either spring.
  • In series, each spring carries the whole load, and their extensions add: 1/k_eff = 1/k₁ + 1/k₂. The pair is less stiff than either spring.

Two identical springs of force constant k give 2k in parallel and k/2 in series, so the same load stretches the series pair four times as far.

Exam-style question 1

Two springs in series

7 marks

Examination question

Springs A and B have force constants 6.0 N m⁻¹ and 3.0 N m⁻¹. They are joined end to end, one end is fixed, and a force of 0.60 N pulls on the other end. Both obey Hooke’s law.

(a) Find the extension of each spring. [2 marks]

(b) Find the force constant of the pair. [3 marks]

(c) Find the total work done in stretching the pair. [2 marks]

Work through (a), (b) and (c)

Unit: m
Unit: m
Unit: N m⁻¹
Unit: J

Show solution step by step

  1. (a) Extensions

    2 marks

    Method

    Divide the force by each force constant.

    Reason

    Springs in series each carry the full force.

    Working

    xA = 0.60/6.0 = 0.10 m and xB = 0.60/3.0 = 0.20 m

  2. (b) Force constant of the pair

    3 marks

    Method

    Add the extensions, then divide the force by the total.

    Reason

    The pair stretches by both extensions together under the same force.

    Working

    x = 0.30 m, so k = 0.60/0.30 = 2.0 N m⁻¹. Check: 1/6.0 + 1/3.0 = 1/2.0.

  3. (c) Work done

    2 marks

    Working

    W = (1/2)Fx = (1/2)(0.60)(0.30) = 0.090 J

Common mistakes

  • Putting the total length of the spring into F = kx instead of the extension.
  • Forgetting to convert centimetres or millimetres to metres before finding k.
  • Using F = kx beyond the limit of proportionality.
  • Mixing up the limit of proportionality with the elastic limit.
  • Using Fx instead of (1/2)Fx for the work done in stretching a spring.

Before you move on

Check your understanding 3

Without looking back: state Hooke’s law and when it is valid. What does x measure? Which is stiffer, two springs in series or the same two springs in parallel?

Show answer

F = kx: the force is proportional to the extension, up to the limit of proportionality. x is the extension from the natural length. The parallel pair is stiffer, because the two springs share the load.

Syllabus and review details

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