The spring is given a compression a and released. The block of mass m again comes to rest when the spring extends by b. During this process, the work done by the spring on the block is W. If $μ$ be the coefficient of friction between the block and the surface, then

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

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

$μ = \frac{k (a - b)}{2 mg}$

$μ = \frac{k (a + b)}{2 mg}$

answer is B, C.

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Detailed Solution

Work done by spring on block is

$W = \frac{1}{2} {ka}^{2} - \frac{1}{2} {kb}^{2} = \frac{1}{2} k (a^{2} - b^{2})$

According to MWET, we have $W_{nc} = ΔU + ΔK$

Since $ΔK = 0$

$\begin{array}{l} \Rightarrow - μmg (a + b) = \frac{1}{2} k (b^{2} - a^{2}) \\ \Rightarrow μmg (a + b) = \frac{1}{2} k (a^{2} - b^{2}) \\ \Rightarrow μ = \frac{k (a - b)}{2 mg} \end{array}$

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