Question

A 1 kg of block situated on a rough incline is connected to a spring of spring constant 100n/m. the block is released from rest with the spring in unstretched position. the block moves 10 cm down the incline plane before coming to rest. find the coefficient of friction between the block and incline. assume that the spring has a negligible mass and the pulley in frictionless

Answer 1

Given, Force constant (K) = 100 N/m angle of inclinations = 37° break in components of weight (mg ) in equilibrium , Normal reaction of the inclined plane ( N) = mgcos∅ 
we know, friction = uN where u is coefficient of friction .friction = umgcos∅ .hence, net Force acting along inclined plane = mgsin∅ - fr = mgsin∅ - umgcos∅ 
in equilibrium condition , work done = PE of stretched spring net force x displacement = 1/2kx² [ use formula , PE = 1/2 Kx² ] (mgsin∅ - umgcos∅)× x = 1/2Kx² here, m = 1 Kg K = 100 N/m x = 10cm g = 10 m/s² ∅ = 37° 
u = (2mgsin37° - kx)/2mgcos37°= ( 2×1×10×3/5 - -100×0.1)/2×1×10×4/5 = (12 - 10)/16 = 2/16 = 1/8 u = 1/8 hence ,coefficient of friction = 1/8