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Elastic strain energyEdexcel A-Level Physics: Revision notes

Section 1

Work done in stretching

To stretch a spring or wire, a force must do work. If the force were constant, work = force × distance. A stretching force is not constant, because it rises from zero as the extension increases. Within the elastic limit this work is stored in the material as elastic strain energy and is recovered when the force is removed.

The work done equals the area under the force–extension graph.

Key termselastic strain energywork done

Section 2

Eel = ½Fx

For a material obeying Hooke's law, the graph is a straight line through the origin, so the area is a triangle with base x and height F:

Eel=12FxE_{el} = \tfrac{1}{2}Fx

The average force during the stretch is F/2, which is why the energy is half of F × x. Since F = kx, this is also ½kx². Doubling the extension doubles both F and x, so the energy rises four times.

Worked example: k = 250 N m⁻¹, x = 0.060 m. F = 15 N, so E = ½ × 15 × 0.060 = 0.45 J. At x = 0.120 m, E = ½ × 30 × 0.120 = 1.8 J.

Key termsEel
Common mistake

Using E = Fx. This forgets that the force starts at zero, so the average force is only F/2.

Section 3

Non-linear graphs: estimating the area

When the graph is not a straight line, ½Fx does not apply, so the energy must be found from the area under the curve. Two methods:

  • Counting squares: find the energy represented by one square, and count whole and part squares.
  • Trapezia (or strips): divide the area into strips of equal width h. Each trapezium has area ½(F₁ + F₂)h; add them up.

Worked example: a polymer gives 0, 3.0, 5.0 N at extensions 0, 2.0, 4.0 mm. Trapezia: ½(0 + 3.0) × 2.0 + ½(3.0 + 5.0) × 2.0 = 3.0 + 8.0 = 11 N mm = 1.1 × 10⁻² J. Convert N mm to J by multiplying by 10⁻³.

For a curve whose gradient decreases, the chords lie below the curve, so trapezia underestimate the area.

Key termstrapezium rulearea under a graph
Exam tip

Show the area of each strip separately. Method marks are given even if the final estimate is slightly out.

Section 4

Energy changes

Elastic strain energy can be transferred to other forms when the force is removed. For a spring launching a trolley on a frictionless track, all the stored energy can become kinetic energy:

12Fx=12mv2\tfrac{1}{2}Fx = \tfrac{1}{2}mv^2

In real systems some energy is transferred to thermal energy and sound, or to the kinetic energy of other parts, so the speed is less than this ideal value. If the energy is multiplied by a factor n, the speed is multiplied by √n, since v ∝ √E.

Key termskinetic energyenergy transfer

Must Know

  • Elastic strain energy = area under the force–extension graph
  • Eel = ½Fx (equals ½kx² when F = kx), a triangle for a linear graph
  • Doubling the extension gives four times the energy
  • Non-linear graph: count squares or use trapezia
  • Speed from stored energy: ½mv² = Eel

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Elastic strain energy

  1. A spring of stiffness 250 N m⁻¹ obeys Hooke's law. It is extended by 0.060 m.
    The spring is attached to a trolley of mass 0.30 kg on a frictionless horizontal track. When released from this extension, all of the stored energy becomes kinetic energy of the trolley. Calculate the maximum speed of the trolley.2 marks
  2. A metal wire obeys Hooke's law up to a force of 40 N, at which its extension is 2.0 mm.
    A student calculates the work done in stretching the wire to 40 N as 40 N × 2.0 mm = 0.080 J. Explain why this is incorrect and give the correct value.2 marks
  3. A strip of elastic polymer is stretched within its elastic limit, and the force is measured at 2.0 mm intervals of extension. The readings are: 0 N at 0 mm, 3.0 N at 2.0 mm, 5.0 N at 4.0 mm, 6.0 N at 6.0 mm and 6.5 N at 8.0 mm. The graph of force against extension is a curve whose gradient decreases as the extension increases.
    Estimate the elastic energy stored in the strip when its extension is 4.0 mm.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).