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2.1.2Energy

Elastic Potential Energy — Oxford AQA International GCSE Combined Science - Physics Revision Notes & Topical Questions

5 min read 8 parts Lesson 3 of 12

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Part 1 of 6Storing energy in an elastic object

Elastic potential store
  • Elastic potential store An object that changes shape when forces act on it (stretching, compressing or bending) stores energy in its elastic potential store.
  • Springs, elastic bands, bows and trampolines all work this way, and the more the shape is changed, the more energy is stored.
Work is done
  • Work is done Stretching a spring needs at least two forces acting in opposite directions, otherwise the whole spring would simply move instead of changing shape.
  • The work done on the spring is transferred to its elastic potential store.
Releasing the energy
  • Releasing the energy When the spring is let go, energy is transferred away from this store, e.g. to the kinetic store of an object such as an arrow or a toy car.
  • In an ideal case all the stored energy is transferred usefully, but in practice some is dissipated by heating and sound.

Part 2 of 6The equation

QuantitySymbolUnit
Elastic potential energyEejoules, J
Spring constantknewtons per metre, N/m
Extensionemetres, m
EquationEe = ½ × k × e2Valid only up to the limit of proportionality

Part 3 of 6Calculating the energy

1
Find the extension: new length minus original length (e.g. 0.230 m − 0.200 m = 0.030 m).
2
Convert every length into metres before calculating; 3.0 cm = 0.030 m.
3
Write the equation Ee = ½ × k × e2 and substitute, e.g. k = 80 N/m: Ee = ½ × 80 × (0.030)2.
4
Square the extension first: (0.030)2 = 0.0009. Then Ee = ½ × 80 × 0.0009 = 0.036 J.

Part 4 of 6Worked examples

QuestionWorkingAnswer
A spring with k = 200 N/m is stretched by 0.10 m. Find the energy stored.Ee = ½ × 200 × (0.10)2 = ½ × 200 × 0.011.0 J
A spring of original length 12 cm is stretched to 15 cm and k = 50 N/m.e = 3 cm = 0.03 m; Ee = ½ × 50 × (0.03)20.0225 J
A spring stores 0.40 J when stretched by 0.04 m. Find k.k = 2Ee ÷ e2 = (2 × 0.40) ÷ (0.04)2500 N/m
A catapult band stores 2.5 J and all of it is transferred to a 0.050 kg stone. Find the speed.½mv2 = 2.5, so v2 = 2 × 2.5 ÷ 0.050 = 10010 m/s

Part 5 of 6Limit of proportionality

1Small forces
  • Small forces Extension is directly proportional to force, so the spring obeys Hooke's law, F = k × e.
2Limit reached
  • Limit reached The graph of force against extension stops being a straight line and begins to curve.
3Beyond the limit
  • Beyond the limit The equation for elastic energy no longer gives the correct value and the spring may be permanently deformed,
    • so it will not return to its original length.

Part 6 of 6Everyday applications

Elastic potential energy in use
Archery
  • Archery Drawing the bow stores energy in the bent limbs;
    • releasing the string transfers it to the kinetic store of the arrow.
Vehicle suspension
  • Vehicle suspension Springs compress over bumps, storing energy briefly and then releasing it, which smooths the ride.
Trampolines
  • Trampolines The springs and fabric stretch as a person lands,
    • then transfer the energy back to the person's kinetic and gravitational potential stores.
Exam tip
  • Exam tip Link the stored energy to the work done in stretching: for a spring obeying Hooke's law, the energy equals the area under the force-extension graph.

Key terms

TermMeaning
Elastic potential energyEnergy stored in an object when it is stretched, compressed or bent by forces.
Spring constantThe force needed per unit extension of a spring; a stiffer spring has a larger value.
ExtensionThe increase in length of a spring from its original length.
Limit of proportionalityThe point beyond which extension is no longer proportional to the force applied.
Elastic deformationA change of shape that is reversed when the forces are removed.
Hooke's lawThe extension of a spring is directly proportional to the force applied, up to the limit of proportionality.

Avoid these mistakes

1
Using the total length of the spring instead of its extension.
2
Forgetting to convert cm or mm to metres, which gives an answer far too large.
3
Squaring the whole product, not just the extension.
4
Using the equation for a spring stretched beyond its limit of proportionality.
5
Forgetting the ½ in the equation, which doubles the answer.

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