Hook's Law

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Question

If the one end of a wire is fixed with a rigid support and the other end is stretched by a force of 10N, then the increase in length is 0.5mm. The ratio of the energy of the wire and the work done in displacing it through 0.5 mm by the same force

Moderate

Solution

work done in stretching the wire,
$W = \frac{1}{2} F l = \frac{1}{2} \times 10 \times 0.5 \times 10^{- 3} = 2.5 \times 10^{- 3}$
Work done to displace it through 0.5 mm
$W = F \times l = 5 \times 10^{- 3}$
So the required ratio is 1:2

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