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The frictional force $f_k$ acting on the particle is given by $f_k = \mu_k(x) N$. Since the motion is on a horizontal surface, the normal force $N = Mg$. Thus, $f_k(x) = (\mu_0 - \alpha x) Mg$.
We are given that $\mu_k(L) = 0$. Substituting this into the expression: $$ \mu_0 - \alpha L = 0 \implies \alpha = \frac{\mu_0}{L} $$
Work done by friction is the integral of force over the displacement. Since friction opposes motion, $W = -\int_{0}^{L} f_k(x) dx$: $$ W = -\int_{0}^{L} (\mu_0 - \alpha x) Mg \, dx $$ $$ W = -Mg \left[ \mu_0 x - \frac{\alpha x^2}{2} \right]_{0}^{L} $$
Substitute $\alpha = \frac{\mu_0}{L}$ into the integral result: $$ W = -Mg \left( \mu_0 L - \frac{\mu_0}{L} \cdot \frac{L^2}{2} \right) $$ $$ W = -Mg \left( \mu_0 L - \frac{\mu_0 L}{2} \right) = -\frac{1}{2} \mu_0 MgL $$ The magnitude of work done is $\frac{1}{2} \mu_0 MgL$. Comparing this with $n \mu_0 MgL$, we get $n = \frac{1}{2}$.
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