Q: A small mass slides down an inclined plane of inclination θ with the horizontal. The coefficient of friction is μ = μ0x where x is the distance through which the mass slides down and μ0 a positive constant. Then the distance covered by the mass before it stops is
(a) $ \displaystyle \frac{2}{\mu_0}tan\theta $
(b) $\displaystyle \frac{4}{\mu_0}tan\theta $
(c) $\displaystyle \frac{1}{2\mu_0}tan\theta $
(d) $ \displaystyle \frac{1}{\mu_0}tan\theta $
Ans: (a)
Sol: Work done by force m g sinθ = Work done against friction
$\large \int_{0}^{x}mgsin\theta dx = \int_{0}^{x}\mu mgcos\theta dx$
$\large \int_{0}^{x}mgsin\theta dx = \int_{0}^{x}\mu_0 x mgcos\theta dx$
$\large mgsin\theta x = \mu_0 mgcos\theta (x^2/2)$
$\large sin\theta = \mu_0 cos\theta (x/2)$
$ \displaystyle x = \frac{2}{\mu_0}tan\theta $