A small blockofmass 'm' is released from rest at the top of a rough inclined plane as shown The coefficient of friction between the block and inclined plane is `'mu'` Using work-energy theorem, find the speed of block as it passes the lowest point.

Explanation

The FBD of the block is shown.

initial speed is zero final speed is v ,brgt By work-energy therorm `1/2mv^(2)=0=W_("Total")` We shall find the work done by each force
(1) `W_(N)=0`
(2) `W_(mgcos theta) =0`
(3) `W_(mgsin theta)=mgsinthetaxxSxxcos0^(@)=mgsinthetaxxS-mg((H)/(S))xxS=mgH(becausesintheta=H/S)`
(4) `W_(fk)=muNxxSxxcos180^(@)=-mumgcosthetaxxS=-mumg((L)/(S))xxS=-mumgL(becausecos theta=L/S)`
`therefore W_("Total)=mgH-mumgL`
`therefore1/2mv^(2)=mgH-mumgLor v=sqrt(2g(H-muL))`