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
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
| Quantity | Symbol | Unit |
|---|---|---|
| Elastic potential energy | Ee | joules, J |
| Spring constant | k | newtons per metre, N/m |
| Extension | e | metres, m |
| Equation | Ee = ½ × k × e2 | Valid 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
| Question | Working | Answer |
|---|---|---|
| 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 × 0.01 | 1.0 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)2 | 0.0225 J |
| A spring stores 0.40 J when stretched by 0.04 m. Find k. | k = 2Ee ÷ e2 = (2 × 0.40) ÷ (0.04)2 | 500 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 = 100 | 10 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
| Term | Meaning |
|---|---|
| Elastic potential energy | Energy stored in an object when it is stretched, compressed or bent by forces. |
| Spring constant | The force needed per unit extension of a spring; a stiffer spring has a larger value. |
| Extension | The increase in length of a spring from its original length. |
| Limit of proportionality | The point beyond which extension is no longer proportional to the force applied. |
| Elastic deformation | A change of shape that is reversed when the forces are removed. |
| Hooke's law | The 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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