A block of mass `0.18kg` is attached to a spring of force constant `2Nm^-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 figure. The block slides a distance of `0.06m` and comes to rest for the first time. The initial velocity of the block in `ms^-1` is `V=N//10`. Then N is

Answer

4

Explanation

`implies 1/2 m u^2 = mu m g xx 0.06 + 1/2 K x^2`
`1/2 xx 0.18 u^2 = 0.1 xx 0.18 xx 10 xx 0.06 + 1/2 xx 2 xx (0.06)^2`
`u = 0.4 ms^(-1), " " 0.4 = N/10 implies N = 4`