AQA · GCSE · Combined Science · Higher

Spring extension from elastic energy

Q03.2 5 marks

When the child is at position A, each trampoline spring is stretched by 0.056 m

The elastic potential energy of each spring is 4.9 J

When the child is at position B, the elastic potential energy of each spring increases to 8.1 J

Calculate the extension of each spring when the child is at position B.

Use the Physics Equations Sheet.

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Marking points

1 Use $E_e = \frac{1}{2}ke^2$ with $E_e=4.9\ \mathrm{J}$ and $e=0.056\ \mathrm{m}$ for position A.
2 Calculate the spring constant as $3125\ \mathrm{N/m}$ .
3 Use $E_e = \frac{1}{2}ke^2$ again for position B with $E_e=8.1\ \mathrm{J}$ and the calculated spring constant.
4 Rearrange correctly to make the extension the subject.
5 Give the extension as $0.072\ \mathrm{m}$ .

Full-mark answer

Use $E_e=\frac{1}{2}ke^2$ . At position A, $4.9=0.5\times k\times0.056^2$ , so $k=3125\ \mathrm{N/m}$ . At position B, $8.1=0.5\times3125\times e^2$ , so $e=\sqrt{\frac{2\times8.1}{3125}}=0.072\ \mathrm{m}$ .

Why this answer loses marks

$e^2=0.005184$ , so the extension is $0.005184\ \mathrm{m}$ .

This calculates $e^2$ but does not take the square root to find $e$ .

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