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`.

Options

A. `0.5, 1000 N/m`
B. `0.5, 500 N/m`
C. `0.2, 500 N/m`
D. `0.2, 1000 N/m`

Answer

`(1)`

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`
`20 sin 37^(@)(5) - mu xx 20 cos 37^(@) xx 5 -1/2 K(0.2)^(2) = 0`
`80mu + 0.02K = 60 rightarrow (1)`
For the upward motion of the body
`-mg sin theta l_(2) + (f xx l_(2)) + 1/2 Kx^(2) = Delta KE `
`-2 xx 10 sin 37^(@) xx 1 - mu xx 20 cos 37^(@) xx 1 + 1/2 K (0.2)^(2) = 0`
`16mu - 0.02 K = - 12 rightarrow (2)`
Adding equations (1) and (2) , we get `96 mu = 48 Rightarrow mu = 0.5`
Now , use the value of `mu` in equation (1), we get `K = 1000N/m`