A 1 kg block situated on a rough incline isconnected to a springconstant 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.

A 1 kg block situated on a rough incline isconnected to a springconsta

1.	A 1 kg block situated on a rough incline isconnected to a springconstant 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.
Answer» Solution :Massof 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`