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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 $n\mu_0MgL$, where g is the acceleration due to gravity. The value of n is:
Detailed Solution
$\mu_k=\mu_0-\alpha x$ and $\mu_k(L)=0 \Rightarrow \alpha=\frac{\mu_0}{L}$. Work done by kinetic friction $=-\int_0^L(\mu_0-\alpha x)mg\,dx=-\left[\mu_0mgL-\alpha mg\frac{L^2}{2}\right]=-\left[\mu_0mgL-\frac{\mu_0}{L}mg\frac{L^2}{2}\right]=-\frac{\mu_0mgL}{2}$. Comparing the magnitude with $n\mu_0mgL$: $n=\frac{1}{2}$.