2.3.6Forces, Space and Radioactivity
Calculating Work Done in Stretching — WJEC GCSE Science (Double Award) - Physics Revision Notes & Topical Questions
5 min read 6 parts Lesson 18 of 37
Part 1 of 4Work done on a spring (HT only)
Energy is transferred
- Energy is transferred Stretching a spring needs a force to act through a distance, so work is done on the spring.
- This work is stored as elastic potential energy in the stretched spring, so the work done equals the energy stored.
The equation
- The equation W = ½ × F × x,
- where W is the work done or energy stored in joules (J), F is the final stretching force in newtons (N) and x is the extension in metres (m).
- The factor of one half appears because the force grows steadily from zero to F while the spring stretches.
When it applies
- When it applies The equation works only while the spring stays within its limit of proportionality, where force is directly proportional to extension.
Part 2 of 4Using the force-extension graph (HT only)
1Straight line from the origin
- Straight line from the origin Within the limit of proportionality, a graph of force against extension is a straight line through (0, 0).
2Area under the line is a triangle
- Area under the line is a triangle The shaded area between the line, the extension axis and a vertical line at the final extension is a right-angled triangle.
3Triangle area gives the work
- Triangle area gives the work Area = ½ × base × height = ½ × extension × force, which is exactly W = ½ × F × x.
4Same idea as the spring constant
- Same idea as the spring constant Because F = k × x, the work can also be written W = ½ × k × x2, which is useful when the force is not given.
Part 3 of 4Worked example (HT only)
1
Question: a 250 g mass hangs from a spring, which stretches from 10.0 cm to 11.4 cm. Calculate the elastic energy stored. Take g = 10 N/kg.
2
Extension: x = 11.4 cm − 10.0 cm = 1.4 cm, and 1.4 ÷ 100 = 0.014 m.
3
Force: the stretching force is the weight, F = m × g = 0.250 kg × 10 N/kg = 2.5 N.
4
Choose the equation: W = ½ × F × x.
5
Substitute: W = ½ × 2.5 N × 0.014 m = 0.0175 J.
6
State the answer with its unit: 0.0175 J (or 1.75 × 10−2 J) of elastic potential energy is stored.
Part 4 of 4Quantities and units
| Quantity | Symbol | Unit | Conversion to watch |
|---|---|---|---|
| Work done / elastic energy | W | joule, J | 1 J = 1 N m |
| Force | F | newton, N | Weight = m × g with m in kg |
| Extension | x | metre, m | Divide cm by 100 and mm by 1000 |
| Spring constant | k | N/m | Gradient of the force-extension graph |
Key terms
| Term | Meaning |
|---|---|
| Work done | The energy transferred when a force moves through a distance. |
| Elastic potential energy | The energy stored in a stretched or compressed object, such as a spring, that can be released when it returns to its shape. |
| Extension | The increase in length of a spring from its original length. |
| Limit of proportionality | The point beyond which force is no longer directly proportional to extension. |
| Area under a graph | For a force-extension graph, the area equals the work done in stretching. |
Avoid these mistakes
1
Leaving out the ½ and using W = F × x. That equation gives the work for a constant force, not for a spring where the force rises from zero.
2
Using the length of the stretched spring instead of the extension. Subtract the original length first.
3
Not converting centimetres to metres, which makes the answer 100 times too large (or 10 000 times if you square).
4
Using the equation for a spring stretched past its limit of proportionality, where the graph is no longer a straight line and the area is not a simple triangle.
5
Forgetting to find the force from the mass first: the mass in kilograms must be multiplied by g to give newtons.
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