$A$ block of mass $0.18$ kg is attached to a spring of force-constant $2 N / m$ . 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 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 $m / s$ is $V = \frac{N}{10}$ . Then $N$ is:

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Solution

The correct option is A 4
applying work energy theorem
$- \frac{1}{2} k x^{2}$ $-$ $μ m g x$ $=$ $\frac{1}{2} m v^{2}$
$v =$ $\frac{4}{10}$
N $=$ 4

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The correct option is A 4applying work energy theorem−12kx2 − μmgx = 12mv2 v= 410N = 4

The correct option is A 4
applying work energy theorem
$- \frac{1}{2} k x^{2}$ $-$ $μ m g x$ $=$ $\frac{1}{2} m v^{2}$
$v =$ $\frac{4}{10}$
N $=$ 4