A block of mass 0.18 kg is attached to a spring of force constant $2 N m^{- 1}$ .The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is unstretched. An impulse is given to the block as shown in Fig. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block in $m s^{- 1}$ is $v = \frac{N}{10}$ .Then N is

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answer is 4.

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Detailed Solution

$\Rightarrow \frac{1}{2} m u^{2} = μ m g \times 0.06 + \frac{1}{2} K x^{2}$

$\frac{1}{2} \times 0.18 u^{2} = 0.1 \times 1.18 \times 10 \times 0.06 + \frac{1}{2} \times 2 \times {(0.06)}^{2}$

$u = 0.4 m s^{- 1}$

$0.4 = \frac{N}{10} \Rightarrow N = 4$

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