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,

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


1 Answer 1


Use conservation of energy.

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.

  • $\begingroup$ I tried using conservation of energy I don't think I need to account for work done by friction because the velocity I calculated already accounts for it. I tried equating, KE and PE but it give me value of any variable. Also, we have to calculate displacement after it has already covered distance 'l'. $\endgroup$ Oct 27, 2016 at 22:35
  • $\begingroup$ You asked for a hint. I think conservation of energy gives the easiest solution. Can you show your calculation using KE and PE? What went wrong with that calculation? $\endgroup$ Oct 27, 2016 at 22:43
  • $\begingroup$ According to your first statement, I should try to equate KE and the 'loss' of PE. But if I take the surface to be zero, KE=PE. I had value of velocity so I substituted it to find KE, then I equated KE and PE $\endgroup$ Oct 27, 2016 at 22:51
  • $\begingroup$ $\frac12 mv^2 = mgh - W$ 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 $2gl(\sin\theta-\mu\cos\theta)=2\mu gL$. $\endgroup$ Oct 27, 2016 at 22:59

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