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A spring of force constant K is first stretched by distance a form its natural length and then further by distance b . The work done in stretching the part b is

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$\frac{1}{2} k a (a - b)$

$\frac{1}{2} k b (a - b)$

$\frac{1}{2} k b (2 a + b)$

$\frac{1}{2} k a (a + b)$

answer is D.

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Detailed Solution

$w = \frac{1}{2} k (x_{2}^{2} - x_{1}^{2})$

$w = \frac{1}{2} k [(a + b)^{2} - a^{2}]$

$w = \frac{1}{2} k [a^{2} + b^{2} + 2 a b - a^{2}]$

$= \frac{1}{2} k b [2 a + b]$

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