Q: A small mass slides down an inclined plane of inclination θ with the horizontal. The coefficient of friction is μ = μ_{0}*x* where *x* is the distance through which the mass slides down and μ_{0} 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 $