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 force is `0.1` insitially its block is at rest and the block as spring is an streched , As impalse is given to the block as shown in the figure . The block sides a distance of ` 0.06` in and comes to the first time . The initial velocity of the for blocks is mis `V = N 10` then `N` is .

Answer

D

Explanation

Let `v` be the speed of the block just after impulse At B , the block comes to rest . Therefore


Loss in K.E. of the block = Gain in P.E. of the spring
`+` Work done against friction
`(1)/(2) mnu^(2) = (1)/(2) kx^(2) + mu mg . x `
`:. nu = sqrt((k)/(m) x^(2) + mu gx ) .
nu = sqrt((2)/(0.18) xx 0.06 xx 0.06 + 0.1 xx 10 xx 0.06 ))
`:. nu = (4)/(10) :. N + 4 `