As is clear from figure,
`R=mgcostheta, F=muR=mu mg costheta`
Net force on the block doen the incline
`=msintheta-F=mg sing theta - mu mg cos theta = mg (sin theta - mu cos theta )`
distance moved, `x=10cm=0.1m. `
In equilibrium, work done `=` P.E. of stretched spring
`mg(sintheta-mu cos theta )x =(1)/(2)Kx^(2)`
`2mg(sintheta-mu cos theta)x=Kx`
`2xx1xx10(sin 37^(@)-mucos 37^(@))=100xx0.1`
`20(0.601-mu.0.798)=10 :. mu=0.126`