Q: A body of mass ‘m’ starts from rest from the top of a rough inclined plane of inclination ‘θ’ and length ‘l’. The velocity ‘v’ with which it reaches the bottom of incline if μk is the coefficient of kinetic friction
(a) $ \displaystyle \sqrt{2gl(cos\theta – \mu_k sin\theta)} $
(b) $ \displaystyle \sqrt{2gl(sin\theta + \mu_k cos\theta)} $
(c) $ \displaystyle \sqrt{2gl(cos\theta + \mu_k sin\theta)} $
(d) $ \displaystyle \sqrt{2gl(sin\theta – \mu_k cos\theta)} $
Ans:(d)
Sol: Loss in P.E = Gain in K.E + Work done against friction
$ \displaystyle mg(l sin\theta)= \frac{1}{2}m v^2 + (\mu_k mg cos\theta )l $
$ \displaystyle mg(l sin\theta) – (\mu_k mg cos\theta )l= \frac{1}{2}m v^2 $
$ \displaystyle mgl( sin\theta) – \mu_k cos\theta ) = \frac{1}{2}m v^2 $
$ \displaystyle v = \sqrt{2 g l(sin\theta – \mu_k cos\theta)} $