The time taken for an ice block to slide down on inclined surface of inclination 60^(@) is 1-2 times the time taken by the same ice block to slide down a frictionless inclined plane of the same inclination. Calculate the coefficient of friction between ice and the inclined plane ?

The time taken for an ice block to slide down on inclined surface of i

1.	The time taken for an ice block to slide down on inclined surface of inclination 60^(@) is 1-2 times the time taken by the same ice block to slide down a frictionless inclined plane of the same inclination. Calculate the coefficient of friction between ice and the inclined plane ?
Answer» Solution :The force of friction `mu_(k)mg cos theta` opposes the component of theweight `mg sin theta` when theice BLOCK slides down. Acceeration of ice `=a_(1)=g sin theta-mu_(k)gcostheta` `u=0` For the plane with friction, `s=ut+(1)/(2)at^(2)`, `s=0+(1)/(2)a_(1)t_(1)^(2)` For the frictionless plane, `a_(2)=g sin theta` `s=0+(1)/(2)a_(2)t_(2)^(2)` The distance TRAVELLED s is the same in the two cases `a_(1)t_(1)^(2)=a_(2)t_(2)^(2)` `t_(1)^(2)(gsintheta-mu_(k)gcostheta)=gsinthetat_(2)^(2)` `(t_(1)^(2))/(t_(2)^(2))=(gsintheta)/(gsintheta-mu_(k)gcostheta)` `(1.2)^(2)=(SIN60^(@))/(sin60^(@)-mu_(k)cos60)=(0.866)/(0.866-mu_(k)xx0.5)` `0.5mu_(k)=0.866-(0.866)/(1.44)=(0.265)/(0.5)=0.53`