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Two springs A and B are identical except that A is stiffer than B i.e., $k_{A} > k_{B} .$ If the two springs are stretched by the same force, then

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more work is done on A i.e., $W_{A} > W_{B}$

more work is done on B i.e., $W_{B} > W_{A}$

work done on A and B are equal

work done depends upon the way in which they are stretched

answer is A.

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Detailed Solution

$F = k_{A} x_{A} = k_{B} x_{B}$

$W_{A} = \frac{1}{2} k_{A} {x_{A}}^{2}; W_{B} = \frac{1}{2} k_{B} {x_{B}}^{2}$

$\frac{W_{A}}{W_{B}} = \frac{k_{A} {x_{A}}^{2}}{k_{B} {x_{B}}^{2}} = \frac{k_{A} {k_{B}}^{2}}{k_{B} {k_{A}}^{2}} = \frac{k_{B}}{k_{A}}$

Hence $W_{B} > W_{A},$ So more work is done on B.

Hence ${ME}_{\max} = \frac{1}{2} m {(2 u)}^{2} = 2 {mu}^{2}$

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