Elastic Strain Energy (AQA A Level Physics): Revision Note

Exam code: 7408

Katie M

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Elastic strain energy

  • Work has to be done to stretch a material

  • Before a material reaches its elastic limit (whilst it obeys Hooke's Law), all the work done is stored as elastic strain energy

  • The work done, or the elastic strain energy is the area under the force-extension graph

Comparison of two force–extension graphs. Hooke’s law is a straight line with triangular shaded area and work done equals one-half F delta L. Non-Hooke’s law has a curved graph with the full area underneath shaded and is labelled work done equals area under graph.
Work done is the area under the force-extension graph
  • This is true for whether the material obeys Hooke’s law or not

    • For the region where the material obeys Hooke’s law, the work done is the area of a right-angled triangle under the graph

    • For the region where the material doesn’t obey Hooke’s law, the area is the full region under the graph

      • To calculate this area, split the graph into separate segments and add up the individual areas of each

  • When a material does obey Hooke's law, the elastic strain energy, E can be calculated using:

E = 12F∆L

  • Where:

    • E = elastic strain energy (or work done) (J)

    • F = average force (N)

    • ∆L = extension (m)

  • This equation is equivalent to the area of a right-angled triangle under the force-extension graph

  • Using Hooke's law F = k∆L, the elastic strain energy can also be written as:

E = 12k(∆L)2

  • Where:

    • k = spring constant (N m–1)

Breaking stress

  • As greater force is applied on a material, the stress on it increases

  • The breaking stress is the maximum stress a material can stand before it fractures (breaks)

    • A material with high breaking stress is considered ductile

    • This means it can stretch a long way before breaking

    • A common example of this is copper, as well as being a good electrical conductor, copper is ductile so it is a suitable material for making wires

  • The ultimate tensile stress (UTS) is sometimes also marked on a stress-strain graph

    • This is the maximum stress that the material can withstand

  • The UTS and breaking stress can depend on the condition of the material such as its temperature

    • This is very important for engineers when considering materials for a particular structure

    • The material might need to stand extreme temperatures and loads which are taken into consideration

Stress–strain graph showing stress rising linearly with strain to the ultimate tensile stress (UTS), then falling to the labelled breaking point. A dashed line marks the breaking stress. Axes are stress/Pa and strain.
The ultimate breaking stress is the point on the stress-strain graph of maximum stress. The breaking point is where the material fractures

Worked Example

The graph shows the behaviour of a sample of a metal when it is stretched until it starts to undergo plastic deformation.

Force–extension graph with force in newtons against extension in millimetres. Graph is a straight line from 0 to point X at 11.0 millimetres and 450 newtons, then curve XY to Y at 13.5 millimetres and 600 newtons.

What is the total work done in stretching the sample from zero to 13.5 mm extension?

Simplify the calculation by treating the curve XY as a straight line.

[3]

Answer:

Step 1: Split the graph into two areas

  • Let area A be the right-angled triangle from 0 to X

  • Let area B be the trapezium from X to Y

Step 2: Calculate area A

  • Area A is a right-angled triangle where the area is 0.5 × base × height

Area A = 0.5 × (11 × 10−3) × 450 = 2.475 J [1 mark]

Step 3: Calculate area B

  • Section B is a trapezium, which can be split into a right-angled triangle and a rectangle

Atriangle + Arectangle = (12 × base × height)  + (base × height)

Atriangle = 0.5(13.5 × 10−3 − 11 × 10−3)(600 − 450) = 0.1875 J

Arectangle = (13.5 × 10−3 − 11 × 10−3)(450−0) = 1.125 J

Area B = 0.1875 + 1.125 = 1.3125 J [1 mark]

Step 4: Calculate the total work done

  • The total work done is the sum of both areas

Work done = 2.475 + 1.3125 = 3.79 J (3 s.f.) [1 mark]

Worked Example

A spring is extended with varying forces; the graph below shows the results.

Force–extension graph with force from 0 to 10 newtons against extension from 0 to 50 millimetres. The graph is a straight line from the origin and reaches 8.1 newtons at 40 millimetres.

What is the energy stored in the spring when the extension is 40 mm?

[3]

Answer:

Step 1: Calculate the area under the graph

  • The area is a right-angled triangle where the area is 0.5 × base × height

Area = 0.5 × (40 × 10−3) × 8.1 = 0.16 J [1 mark]

Step 2: Conclusion

  • The energy stored is the area under the graph, so the energy stored is 0.16 J [1 mark]

Spring energy

  • When a vertical spring is extended and contracted, its energy is converted into other forms

  • Although the total energy of the spring will remain constant, it will have changing amounts of:

    • elastic potential energy (EPE)

    • kinetic energy (KE)

    • gravitational potential energy (GPE)

  • When a vertical mass is hanging on a spring and it moves up and down, its energy will convert between the three in various amounts

A mass on a vertical spring at its highest position A, equilibrium position B and lowest position C. A is labelled “GPE = max”, B is labelled “KE = max”, and C is labelled “EPE = max”.
As a mass on a spring moves between A, B and C, its energy is transferred between elastic potential, kinetic and gravitational potential energy
  • At position A:

    • The spring has some EPE since it is slightly compressed

    • Its KE is 0 since it is stationary

    • Its GPE is at a maximum because the mass is at its highest point

  • At position B:

    • The spring has some EPE since it is slightly stretched

    • Its KE is at a maximum as it passes through the equilibrium position at its maximum speed

    • It has some GPE since the mass is still above the ground

  • At position C:

    • The spring has its maximum EPE because it is at its maximum extension

    • Its KE is 0 since it is stationary

    • Its GPE is at a minimum because it is at its lowest point above the Earth's surface

  • For a horizontal mass on a spring system, there is no gravitational potential energy to consider

    • The spring only converts between kinetic and elastic potential energy

Examiner Tips and Tricks

Make sure you remember which position gives the mass its greatest amount of each type of energy, and where each type of energy is zero.

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Katie M

Author: Katie M

Expertise: Curriculum Expert

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.

Tim

Reviewer: Tim

Expertise: Content Creator

Timothy graduated with a first class degree in Mathematics and Physics from the University of Warwick. After working as a postgraduate researcher, Timothy has worked as a content creator for various online revision platforms, creating physics resources for a range of levels and exam boards.