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Hooke's Law

Key idea: Apply F = kx, interpret force–extension graphs, distinguish proportional and elastic limits, and calculate elastic energy.

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Apply Hooke's law within the limit of proportionality.
Core syllabus scope

Hooke’s law, F = kx, is required in H2 Physics 9478. Young’s modulus and detailed material stress–strain behaviour are not required here; they are labelled as extension below.

1. Definitions

Hooke’s law

For a spring or material in its proportional region, the applied force magnitude F is directly proportional to its extension x from natural length:

F = kx

The force constant k measures stiffness and has unit N m⁻¹.

Elastic and plastic behaviour

An elastic deformation disappears when the load is removed. A plastic deformation leaves a permanent change in shape or length.

The limit of proportionality ends the straight-line F–x region. The elastic limit is the greatest load for which the object returns to its original dimensions after unloading. These limits are conceptually different and need not occur at the same point.

2. Key ideas

  • Hooke’s law applies only in the proportional region: F = kx
  • The force constant is the gradient of a force–extension graph: k = (Δ F)/(Δ x)
  • A larger k means a stiffer spring and a smaller extension for the same force.
  • Elastic potential energy is the area under the force–extension graph. For a linear spring loaded from zero: Eₑₗ = (1/2)kx²
  • Extension x is the change from natural length, not the total stretched length.

3. Detailed explanation

Reading a force–extension graph

In the straight region through the origin, a constant gradient gives k. Beyond the proportional limit the graph is curved, so one value of k no longer describes the response.

Do not infer permanent deformation merely because the graph becomes curved. Permanent deformation begins only beyond the elastic limit and is established by unloading behaviour.

Force–extension graph with proportional and elastic limitsForce 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
Scroll diagram horizontally to read all labels.
The proportional limit P ends the straight Hooke’s-law region. The elastic limit E is a different condition: deformation remains reversible only up to E.

Effective force constant

For ideal springs in parallel:

k_eff = k₁ + k₂ + …

For ideal springs in series:

1/k_eff = 1/k₁ + 1/k₂ + …

Parallel springs are stiffer than either spring alone; series springs extend more under the same load.

Elastic energy

Work done by a gradually increasing force equals the area under the F–x graph. For a linear spring this is a triangle:

Eₑₗ = 1/2 Fx = 1/2 kx²

The shortcut (1/2)kx² is valid only while F = kx. For a non-linear graph, determine the area under the actual curve.

Optional extension: Young's modulus

Young’s modulus compares materials using stress divided by strain, E = σ/ε, within the linear elastic region. It is not a required syllabus 9478 outcome for this topic.

4. Common Mistakes

  • Treating k as a constant even beyond the limit of proportionality (it is only constant in the linear region).
  • Mixing up extension Δ x with the total length of the spring.
  • Forgetting unit conversions (e.g. cm → m) and getting k wrong by a factor of 100.

5. Exam Tips

  • If you’re asked for k from a graph, use two far-apart points on the linear region to find the gradient.
  • State the condition clearly: “Hooke’s law holds up to the limit of proportionality.”
  • For energy, use the area under the F–x graph or, in the linear region, E = (1/2)kx².
  • If the graph has a non-zero intercept, do not force it through the origin; quote the linear-region gradient for k and mention possible systematic offset.

6. Worked Examples

Modelled example 1

What is the magnitude of the force exerted on a spring with a spring constant of 65.0 N m⁻¹ when the spring undergoes a 12.3 cm extension?

Core

Problem

Find the force magnitude for k = 65.0 N m⁻¹ and extension x = 12.3 cm.
Study the worked solution

  1. Check the Hooke-law region

    Method

    Use F = kx for the stated spring response.

    Reason

    The force constant describes the proportional region.

    Working

    F = kx

  2. Convert extension

    Method

    x = 0.123 m.

    Reason

    The force constant is expressed per metre.

    Working

    12.3 cm = 12.3 × 10⁻² m = 0.123 m

  3. Calculate force

    Method

    F = 8.00 N.

    Reason

    Multiply stiffness by extension.

    Working

    F = (65.0)(0.123) = 8.00 N

Guided practice 2

Extension of a vertical spring

About 4 min

Problem

A 7.10 kg bag hangs at rest from a spring with k = 85.0 N m⁻¹. Find the extension. Take g = 9.81 m s⁻².

Try this before viewing the solution

Unit: m

Hints

Hint 1: start with equilibrium
Draw the bag’s forces and set upward kx equal to downward mg.
View solution step by step

  1. Use vertical equilibrium

    Method

    kx = mg.

    Reason

    The stationary bag has zero resultant force.

    Working

    kx-mg = 0

  2. Rearrange and calculate

    Method

    x = 0.819 m.

    Reason

    Divide the bag’s weight by the spring force constant.

    Working

    x = mg/k = (7.10)(9.81)/85.0 = 0.819 m

Common misconception 3

Extension is not total length

Find and correct the mistake

Learner claim

A spring has natural length 0.25 m and stretches to 0.31 m. Its force constant is 100 N m⁻¹. A learner substitutes x = 0.31 m into F = kx. Diagnose the method and find the force.

Try this before viewing the solution

Unit: N

View solution step by step

  1. Identify the required displacement

    Method

    x is extension from natural length.

    Reason

    Hooke’s law does not use the spring’s total length.

    Working

    x = L-L₀

  2. Calculate extension

    Method

    x = 0.060 m.

    Reason

    Subtract the unloaded length from the loaded length.

    Working

    x = 0.31-0.25 = 0.060 m

  3. Calculate force

    Method

    F = 6.0 N.

    Reason

    Use the extension in the proportional-region relation.

    Working

    F = (100)(0.060) = 6.0 N

Examiner practice 4

A spring A of force constant 6.0 Nm-1 is connected in series with a spring B of force constant 3.0 Nm-1. One end of the combination is securely anchored and a force of 0.60 N is applied to the other end.

7 marks

Examination question

For the stated series combination, find (a) each extension, (b) the effective force constant, and (c) total elastic potential energy. [7 marks]

Try this before viewing the solution

Unit: m
Unit: m
Unit: N m^-1
Unit: J

View solution step by step

  1. Find both extensions

    2 marks

    Method

    x_A = 0.100 m and x_B = 0.200 m.

    Reason

    Series springs carry the same tensile force.

    Working

    x_A = 0.60/6.0 = 0.100 m, x_B = 0.60/3.0 = 0.200 m

  2. Find total extension

    1 mark

    Method

    xₜₒₜₐₗ = 0.300 m.

    Reason

    Series extensions add.

    Working

    0.100 + 0.200 = 0.300 m

  3. Find effective force constant

    2 marks

    Method

    k_eff = 2.0 N m⁻¹.

    Reason

    The combination produces the total extension under the common force.

    Working

    k_eff = 0.60/0.300 = 2.0 N m⁻¹

  4. Calculate stored energy

    2 marks

    Method

    Eₑₗ = 0.090 J.

    Reason

    The linear combination stores the triangular force–extension area.

    Working

    Eₑₗ = (1/2)k_effxₜₒₜₐₗ² = (1/2)(2.0)(0.300)² = 0.090 J

Challenge 5

Force–extension point and stored energy

Minimal support

Independent transfer

A linear force–extension graph passes through x = 0.080 m at F = 16 N. Find the force constant and the elastic energy stored at this extension.

Try this before viewing the solution

Unit: N m^-1
Unit: J

Hints

Hint 1: read gradient and area
For a straight line through the origin, k = F/x and E = (1/2)Fx.
View solution step by step

  1. Infer stiffness

    Method

    k = 200 N m⁻¹.

    Reason

    The gradient of the linear force–extension graph is the force constant.

    Working

    k = 16/0.080 = 200 N m⁻¹

  2. Calculate graph area

    Method

    Eₑₗ = 0.64 J.

    Reason

    The area under the straight line is a triangle.

    Working

    Eₑₗ = (1/2)Fx = (1/2)(16)(0.080) = 0.64 J

7. Mind Stretchers

Mind stretcher 1: ChallengeExtension

Two springs have constants k and 2k. If the same energy E is stored in each spring, which spring stretches more? Use E = (1/2)kx².

Show answer

For fixed energy, x = square root of (2E/k). Therefore the spring with force constant k stretches square root of 2 times as far as the spring with force constant 2k.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027