A block of mass m is moving with a speed v on a horizontal rough surface and collides with a horizontally mounted spring of spring constant k as shown in the figure. The coefficient of friction between the block and the floor is $μ$ . The maximum compression of the spring is

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$- \frac{μmg}{k} - \frac{1}{k} \sqrt{{(μmg)}^{2} - {mku}^{2}}$

$\frac{μmg}{k} + \frac{1}{k} \sqrt{{(μmg)}^{2} + {mku}^{2}}$

$\frac{μmg}{k} + \frac{1}{k} \sqrt{{(μmg)}^{2} - {mku}^{2}}$

$- \frac{μmg}{k} + \frac{1}{k} \sqrt{{(μmg)}^{2} + {mku}^{2}}$

answer is A.

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Detailed Solution

Let x be the compression of spring.
Loss in K.E of the block = gain in elastic P.E of the spring + Work done against friction
$\frac{1}{2} {mv}^{2} - 0 = \frac{1}{2} {kx}^{2} + μmgx \Rightarrow {kx}^{2} + 2 μmgx - {mv}^{2} = 0$
by solving quadratic expression,
$x = \frac{- μmg}{k} + \frac{1}{k} \sqrt{{(μmg)}^{2} + {mkv}^{2}}$