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 μ_{k} 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)} $