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Elastic strain energyAQA A-Level Physics: Revision notes

Section 1

Work done in stretching

When a force stretches a wire or spring, it does work, and while the material stays elastic this work is stored as elastic strain energy. If the force is released, the energy is transferred back, for example to the kinetic energy of a launched object.

The work done equals the area under a force–extension graph. For a material obeying Hooke's law the graph is a straight line through the origin, so the area is a triangle.

Key termselastic strain energywork done
Exam tip

If the graph is not a straight line, find the energy by counting squares under the curve and scaling by the value of each square.

Section 2

Calculating the energy stored

For a straight-line graph, the area under it is 12FΔL\frac{1}{2} F \Delta L, where FF is the final force and ΔL\Delta L is the extension, so E=12FΔLE = \frac{1}{2} F \Delta L.

Using Hooke's law, F=kΔLF = k \Delta L, this becomes E=12kΔL2E = \frac{1}{2} k \Delta L^2. Doubling the extension therefore quadruples the stored energy.

Worked example: a spring of k=200k = 200 N m⁻¹ is extended by 0.050 m. F=200×0.050=10F = 200 \times 0.050 = 10 N, so E=12×10×0.050=0.25E = \frac{1}{2} \times 10 \times 0.050 = 0.25 J.

Key termsextension
Common mistake

Using E = F ΔL. The force rises from zero to F, so the average force is F / 2 and the factor of ½ is needed.

Section 3

Energy conservation with springs

A stretched or compressed spring can transfer its elastic strain energy to other stores. Ignoring losses, elastic strain energy = kinetic energy for a horizontal launch, and elastic strain energy = kinetic energy + gravitational potential energy if the object rises.

For a ball launched vertically from a spring compressed by ΔL\Delta L, which rises a height hh above the point where it leaves the spring: 12kΔL2=mg(h+ΔL)\frac{1}{2} k \Delta L^2 = m g (h + \Delta L), because the ball also rises through the compression distance while it is pushed.

In real systems some energy is transferred to thermal energy through friction and air resistance, so the observed speed or height is less than predicted.

Key termsconservation of energy
Common mistake

Forgetting that the object also rises while the spring is still pushing it. Include the gravitational potential energy gained over the compression distance.

Section 4

Beyond the elastic limit

Beyond the elastic limit, a metal deforms plastically: layers of atoms slip past each other and the material does not return to its original length.

On unloading, the line on the graph falls parallel to the original straight section but ends at a permanent extension. The area between the loading and unloading lines is the energy transferred to the material as thermal energy, which is why a metal wire or paper clip warms when bent back and forth.

Key termselastic limitplastic deformation

Section 5

Energy and ethical transport design

Engineers use elastic strain energy to recover energy that would otherwise be wasted, for example in a regenerative braking system that stores kinetic energy in a spring or flywheel and returns it on acceleration.

An evaluation weighs the benefits against the costs:

  • Energy recovered reduces fuel or electricity use, so emissions and air pollution fall
  • Less friction braking reduces brake wear and particulates
  • The added mass of the store increases the energy needed to accelerate
  • Manufacturing and disposal use energy and resources (whole-life cost)
  • No system is 100% efficient, so some energy is always transferred to thermal energy
Key termsregenerative brakingwhole-life cost

Must Know

  • Energy stored = area under the force–extension graph
  • E=12FΔL=12kΔL2E = \frac{1}{2} F \Delta L = \frac{1}{2} k \Delta L^2 for a material obeying Hooke's law
  • Doubling the extension quadruples the energy
  • Elastic strain energy is transferred to kinetic and gravitational potential energy
  • Beyond the elastic limit the unrecovered energy becomes thermal energy
  • Ethical design: weigh energy saved against added mass and manufacturing cost

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Elastic strain energy

  1. A spring obeys Hooke's law within its limit of proportionality. A force of 8.0 N extends the spring by 0.040 m.
    Calculate the additional energy that must be transferred to the spring to increase its extension from 0.040 m to 0.080 m.2 marks
  2. A technician loads a steel wire with increasing weights until it has been stretched beyond its elastic limit, and then removes the weights one by one. After unloading, the wire is found to be permanently longer than at the start and feels slightly warm.
    Explain, in terms of energy, why the wire is slightly warm after it has been unloaded.2 marks
  3. A pinball machine launches a steel ball of mass 0.080 kg using a spring of spring constant 320 N m⁻¹. The spring is compressed by 0.045 m and then released. The ball leaves the spring when the spring reaches its natural length. Take g = 9.81 m s⁻².
    Calculate the speed of the ball as it leaves the spring, assuming the spring energy is transferred entirely to the kinetic energy of the ball.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).