To solve this problem, we need to calculate the total work done by the variable kinetic friction force on a particle moving from \(x = 0\) to \(x = L\) along a horizontal axis.
The kinetic frictional force at any position \(x\) is given by:
\(f_k(x) = \mu_k(x) \cdot N\)
where \(\mu_k(x) = \mu_0 - \alpha x\) is the coefficient of friction, and \(N = Mg\) is the normal force (since the motion is horizontal).
Thus, the frictional force becomes:
\(f_k(x) = (\mu_0 - \alpha x)Mg\)
The work done by friction as the particle moves from \(x = 0\) to \(x = L\) is:
\(W = \int_{0}^{L} f_k(x) \, dx\)
\(= \int_{0}^{L} (\mu_0 - \alpha x)Mg \, dx\)
\(= Mg \int_{0}^{L} (\mu_0 - \alpha x) \, dx\)
Now we integrate:
\(W = Mg \left[ \mu_0 x - \frac{\alpha x^2}{2} \right]_{0}^{L}\)
\(= Mg \left[ \mu_0 L - \frac{\alpha L^2}{2} \right]\)
According to the problem, we have \(\mu_k(L) = 0\), which gives us:
\(\mu_0 - \alpha L = 0\)
\(\alpha = \frac{\mu_0}{L}\)
Substituting \(\alpha\) back into the work expression, we get:
\(W = Mg \left[ \mu_0 L - \frac{\mu_0 L}{2} \right]\)
\(= Mg \left[ \frac{\mu_0 L}{2} \right]\)
\(= \frac{1}{2} \mu_0 Mg L\)
Thus, comparing with \(n \mu_0 Mg L\), we find:
\(n = \frac{1}{2}\)
Therefore, the value of \(n\) is \(\frac{1}{2}\).
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