# Displacement of mass on an inclined plane if it continues sliding

Q. A block of mass m, slides down an inclined plane of length l and coeff. of friction u. With what speed will it reach the bottom? If it further slides on a similar horizontal surface, how far will it go before coming to rest.

My attempt,

$$mg\sin\theta-u(mg\cos\theta)=F_{net}$$ $$a=g\sin\theta-u(g\cos\theta)$$ $$v=\sqrt{2gl(\sin\theta-u\cos\theta}$$

Friction force on incline is $F=\mu N$. Work done against friction is $W=Fl$. So KE at bottom of incline is loss of PE, which is $mgh=mgl\sin\theta$, less the work done against friction.
Use a similar argument to find the distance $L$ which the object slides on the horizontal : KE at bottom of incline = work done on horizontal before object comes to rest.
• $\frac12 mv^2 = mgh - W$ which gives the same expression $v=\sqrt{2gl(\sin\theta-\mu\cos\theta)}$.... On the horizontal part the initial KE is used up against friction : $\frac12mv^2=\mu mgL$ so $2gl(\sin\theta-\mu\cos\theta)=2\mu gL$. – sammy gerbil Oct 27 '16 at 22:59