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GCSE/Combined Science/Edexcel

CP7.2Forces and elasticity: Hooke’s law (F = ke); elastic potential energy = ½ke²; spring constant determination

Notes

Forces and elasticity

Elastic vs inelastic deformation

When a force stretches, compresses or bends an object it deforms.

  • Elastic deformation — the object returns to its original shape when the force is removed (e.g. a spring under a small load).
  • Inelastic deformation — the object stays permanently deformed (e.g. a stretched paper clip).

To deform an object, more than one force must act — pulling at one end alone simply moves it. A spring being stretched has the load pulling down and the support pushing up.

Hooke’s law

For a spring stretched within its limit of proportionality, the extension is directly proportional to the force applied.

F = k × e

  • F = force applied (N)
  • k = spring constant (N/m) — a measure of stiffness
  • e = extension from the natural length (m)

A stiffer spring has a larger k and stretches less for the same force.

A force–extension graph for a spring is a straight line through the origin up to the limit of proportionality, then curves above it (Hooke’s law no longer holds).

Elastic potential energy stored

When a spring is stretched or compressed elastically, work is done on it and energy is stored as elastic potential energy:

E = ½ × k × e²

Units: J. Note the squared — doubling the extension stores four times the energy.

Determining the spring constant (Core Practical link)

  1. Hang a spring from a clamp and measure its natural length with a ruler.
  2. Add masses (e.g. 100 g = 0.98 N) one at a time; measure the new length each time.
  3. Calculate extension = new length − natural length for each load.
  4. Plot force (y-axis) against extension (x-axis).
  5. The gradient of the straight portion = the spring constant k.

Worked example

A spring extends by 0.04 m when a force of 6 N is applied. Calculate (a) the spring constant and (b) the elastic potential energy stored.

(a) k = F ÷ e = 6 ÷ 0.04 = 150 N/m. (b) E = ½ × 150 × 0.04² = ½ × 150 × 0.0016 = 0.12 J.

Edexcel exam tip

Always convert cm to m before calculating with k or E. The most common dropped mark on this topic is "extension in cm". Also, never quote E = ½ × F × e for elastic PE — Edexcel’s mark scheme requires k and e².

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Practice questions

Try each before peeking at the worked solution.

  1. Question 13 marks

    Calculating spring constant

    Edexcel Paper 2F (Foundation)

    A spring extends by 0.05 m when a 10 N weight is hung from it.

    Calculate the spring constant of the spring. State the unit. (3 marks)

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  2. Question 22 marks

    Hooke’s law and limit of proportionality

    Edexcel Paper 2F (Foundation)

    (a) State Hooke’s law. (1 mark)
    (b) State what happens to a force–extension graph beyond the limit of proportionality. (1 mark)

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  3. Question 33 marks

    Elastic potential energy

    Edexcel Paper 2H (Higher)

    A spring with a spring constant of 80 N/m is stretched elastically by 0.15 m.

    Calculate the elastic potential energy stored in the spring. (3 marks)

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Flashcards

CP7.2 — Forces and elasticity: Hooke’s law (F = ke); elastic potential energy = ½ke²; spring constant determination

7-card SR deck for Edexcel GCSE Combined Science — Leaves (batch 5) topic CP7.2

7 cards · spaced repetition (SM-2)