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 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 m//s is `V = N//10`. Then `N` is : .

A block of mass `0.18 kg` is attached to a spring of force-constant `2

1.	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 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 m//s is `V = N//10`. Then `N` is : .
Answer» Correct Answer - `4` Decrease in mechanical energy = work done against friction `(1)/(2) m upsilon^(2) - (1)/(2) kx^(2) = (mu mg ) x` `v = sqrt((2 mugx+k)/(m))` Putting `m = 0.18 kg, x = 0.06 m, k = 2Nm^(-1)`, `mu = 01` we get `upsilon = 0.4 m//s = (4)/(10) m//s` `:. N = 4`.

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