In the figure show below, calculate the coefficient of friction between the block and the inclined plane. Assume that the mass of the block s 1 kg. The block is released from rest and stops by covering the distance of 0.1 m on the plane.

Explanation

Net force on the block = f = `(mgsintheta-mumgcostheta)`
Work done an the block `W=Fx=mgxx(sintheta-mucostheta)`
This energy is equal to `1//2kx^(2)`
`therefore (1)/(2)kx^(cancel2)=mgcancelx(sintheta-mucostheta)`
`(1)/(2)xx10cancel0xx0.1=1xx1cancel0(sin37^(@)-mucos37^(@))`
`mucos37^(@)=sin37^(@)-0.5=0.6018-0.5=0.1018`
`therefore mu=(0.1018)/(0.7986)=0.127`