A blocks of mass 0.18 kg is attached to a spring of force constant $2 {Nm}^{- 1}$ . The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is un-stretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block in ${ms}^{- 1}$ is $V = N / 10 .$ Then N is

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answer is 4.

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Detailed Solution

$\frac{1}{2} {mu}^{2} = μmg \times 0 .06 + \frac{1}{2} {kx}^{2}$
$\Rightarrow \frac{1}{2} \times 0 .18 u^{2} = 0 .1 \times 0.18 \times 10 \times 0 .06 + \frac{1}{2} \times 2 \times (0 .06)^{2}$

$\Rightarrow u = 0 .4 {ms}^{- 1} = \frac{N}{10} \Rightarrow N = 4$

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