A particle of mass M moves along a horizontal | vedclass

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(D) The frictional force $f$ acting on the particle is given by $f = \mu(x) N$, where $N = Mg$ is the normal force.Thus, $f(x) = (\mu_0 - \alpha x) Mg

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$1$/$2$

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(D) The frictional force $f$ acting on the particle is given by $f = \mu(x) N$, where $N = Mg$ is the normal force.Thus, $f(x) = (\mu_0 - \alpha x) Mg$.Given that $\mu(L) = 0$, we have $\mu_0 - \alpha L = 0$, which implies $\alpha = \frac{\mu_0}{L}$.Substituting $\alpha$ into the expression for $f(x)$, we get $f(x) = (\mu_0 - \frac{\mu_0}{L} x) Mg = \mu_0 Mg (1 - \frac{x}{L})$.The work done $W$ by the frictional force is the integral of the force over the displacement from $x=0$ to $x=L$. Since friction opposes motion, $W = -\int_{0}^{L} f(x) dx$.$W = -\int_{0}^{L} \mu_0 Mg (1 - \frac{x}{L}) dx$.$W = -\mu_0 Mg [x - \frac{x^2}{2L}]_{0}^{L}$.$W = -\mu_0 Mg (L - \frac{L^2}{2L}) = -\mu_0 Mg (L - \frac{L}{2}) = -\frac{1}{2} \mu_0 MgL$.The magnitude of the work done is $\frac{1}{2} \mu_0 MgL$. Comparing this with $n\mu_0 MgL$, we find $n = 1/2$.

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