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Question
Identical springs of steel and copper are equally stretched. On which, more work will have to be done?
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Solution
Work Done in stretching a Wire or Spring: In stretching a wire work is done against internal restoring forces. This work is stored in the wire as elastic potential energy or strain energy.
If a force F acts along the length L of the wire of cross-section A and stretches it by x, then work done in stretching a wire is given by W = 1/2F × Δl
As springs of steel and copper are equally stretched. Therefore, for the same force (F),
`W ∝ Δl` ......(i)
Young's modulus `(Y) = F/A xx l/(Δl)`
or `Δl = F/A xx l/Y`
As both springs are identical,
∴ `Δl ∝ 1/Y` ......(ii)
From equations (i) and (ii), we get `W ∝ 1/Y`
∴ `W_(steel)/W_(copper) = Y_(copper)/(Y_(steel)) > 1` ......(As `Y_(steel) > Y_(copper)`)
or `W_(steel) > W_(copper)`
Therefore, more work will be done for stretching copper spring.
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