Question
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:
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Solution
Sol. [Key Concept: Work done against a position-dependent friction (∫F dx)]
Given $\mu_{k}(x) = \mu_{0} - \alpha x$ and $\mu_{k}(L) = 0 = \mu_{0} - \alpha L \Rightarrow \mu_{0} = \alpha L$
Magnitude of work done by frictional force, $W_{f} = \int_{}^{}\ dW_{f}$
where $dW_{f} = f_{k}dx = \mu_{k}Ndx = \mu_{k}Mgdx = \left( \mu_{0} - \alpha x \right)Mgdx$
$$\begin{matrix} & \ \therefore W_{f} = \int_{0}^{L}\mspace{2mu}\mspace{2mu}\mu_{0}Mgdx - \int_{0}^{L}\mspace{2mu}\mspace{2mu}\alpha Mgxdx = \mu_{0}Mg\lbrack x\rbrack_{0}^{L} - \alpha Mg\left\lbrack \frac{x^{2}}{2} \right\rbrack_{0}^{L} \\ & \ \Rightarrow n\mu_{0}MgL = \mu_{0}MgL - \left( \frac{\mu_{0}}{L} \right)Mg\left( \frac{L^{2}}{2} \right) = \frac{\mu_{0}MgL}{2} \\ & \ \Rightarrow n = \frac{1}{2} \end{matrix}$$
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