A 1 kg block situated on a rough incline is connected to a spring constant `100 Nm^(-1)` as shown in figure . The block is released from rest with the spring in the unstretched position . The block moves 10 cm down the incline before coming to rest . Find the coefficient of friction between the block and the incline .Assume that the spring has a negligible mass and the pulley is frictionless.

Explanation

Mass of block m = 1 kg , `g = 10 ms^(-1)`
Spring constant `k = 100 Nm^(-1)`
`theta = 37^(@)` [ and take `sin 37^(@) = 0.6 and cos 37^(@) = 0.8`]
Distance covered by blosk x = 0.1 m
Net force on the block down the incline ,
`F = mg si theta -f`
`= mg sin theta - mu N " " [ :. F = muN] `
` = mg sin theta - mu mg cos theta " " [ :. N = mg cos theta]`

Work done by applying force F for the motion of block ,
`F = Fx = mg(sin theta - mu cos theta)x`
The work done when the block stop stored as potential energy in the spring ,
`W= V = 1/2 kx^(2)`
` :. mgx (sin theta- mu cos theta ) = 1/2 kx^(2)`
` :. sin sin theta - mu cos theta = (1/2kx)/(mg)`
` :. mu cos theta = sin theta -(kx)/(2 mg)`
` :. mu =1/(cos theta ) [ sin theta -(kx)/(2mg)] `
` =1/(cos 37^(@))[ sin 37^(@)-(100xx0.1)/(2xx1xx10)] `
` :. mu =1/(0.8) [0.6 - 0.5]`
` =1/(0.8) [0.1]`
` :. mu = 1/8 = 0.125`