Figure shows a spring fixed at the bottom end of an incline of inclination `37^0`. A small block of mass 2 kg starts slipping down the incline from a point 4.8 m away from the spring. The block compresses the spring by 20 cm, stops momentarily and then rebounds through a distance of 1 m up the incline. Find a. the frictioin coefficient between the plane and the block and b. the spring constant of the spring. Take `g=10 m/s^2`.

Explanation

Applying work energy theorem for downward motion of the body `W = Delta KE`
`mg sin theta(x +d) - f xx l_(1) - (1)/(2) Kx^(2) = Delta KE`
`200 sin 37^(0) (5) - mu xx 20 cos 37^(0) xx 5 -(1)/(2) K(0.2)^(2) = 0`
`80 mu + 0.02 K = 60 rarr (1)`
For the upward motion of the body
`-mg sin theta l_(2) + (f xx l_(2)) + (1)/(2) Kx^(2) = DeltaKE`
`-2xx 10 sin 37^(0) xx 1 - mu xx 20 cos 37^(0) xx 1 + (1)/(2) K(0.2)^(2) = 0`
`16 mu - 0.02 K = -12 rarr (2)`
Adding equations (1) and (2), we get
`96 mu = 48 rArr mu = 0.5`
Now, use the value of `mu` in equation (1),
we get `K = 1000 N//m`.