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) $\dfrac{1}{2}$
Concept. Work by friction $= \displaystyle\int \mu_k(x)Mg\,dx$; $\mu_k(L) = 0$ gives $\alpha = \dfrac{\mu_0}{L}$.
Step 1. Integral.
$$|W| = Mg\int_0^L\left(\mu_0 - \frac{\mu_0x}{L}\right)dx = \frac12\mu_0MgL$$
Step 2. Compare. $n = \dfrac12$ (in magnitude; friction's work is negative).
Answer. $\dfrac12$ (Option D).
Solution by Sreeraj P, M.Sc Physics