An elastic spring of unstretched length L and force constant k is stretched by a small length x. It is further stretched by another small length y. The work done in the second stretching is

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$\frac{1}{2} k y^{2}$

$\frac{1}{2} k (y^{2} - x^{2})$

$\frac{1}{2} k {(x + y)}^{2}$

$\frac{1}{2} k y (2 x + y)$

answer is D.

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Detailed Solution

Potential energy stored in the spring when it is extended by x is

$U_{1} = \frac{1}{2} k x^{2}$

Potential energy stored in the spring when it is further extended by y

is $U_{2} = \frac{1}{2} k {(x + y)}^{2}$

$∴$ work done = $U_{2} - U_{1} = \frac{1}{2} k {(x + y)}^{2} - \frac{1}{2} k x^{2} = \frac{1}{2} k y (2 x + y)$

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