A particle of mass $M$ moves along a horizontal $x$ axis from $x = 0$ to $x = L$. The coefficient of kinetic friction varies as a function of $x$ as $\mu_k(x) = \mu_0 - \alpha x$, where $\mu_0$, $\alpha$ are constants of appropriate dimensions, so that $\mu_k(L) = 0$. The total work done by the frictional force during the motion is $n\mu_0 MgL$, where $g$ is the acceleration due to gravity. The value of $n$ is :
Answer: (D) $\dfrac{1}{2}$
Since $\mu_k(L) = 0$, we get $\mu_0 - \alpha L = 0$, so $\alpha = \dfrac{\mu_0}{L}$.
The friction force at position $x$ is $f = \mu_k(x)\,Mg$. The magnitude of work done by friction is
$$W = \int_0^L \left(\mu_0 - \frac{\mu_0 x}{L}\right) Mg\,dx = Mg\left(\mu_0 L - \frac{\mu_0 L}{2}\right) = \frac{1}{2}\mu_0 MgL$$
(Friction opposes the motion, so this work is negative; its magnitude is $\tfrac{1}{2}\mu_0 MgL$.)
Hence $n = \dfrac{1}{2}$.
Solution by Sreeraj P, M.Sc Physics