RE-NEET 2026 Physics Work and Energy MCQ Question

The work done by the frictional force can be calculated using the formula for work, which is the integral of the frictional force over the distance. The frictional force at position $x$ is given by $F_f = \mu_k(x) Mg = (\mu_0 - \alpha x) Mg$.

To find the total work done by friction from $x=0$ to $x=L$, we compute:

$$W = \int_0^L F_f \, dx = \int_0^L (\mu_0 - \alpha x) Mg \, dx = Mg \left[ \mu_0 x - \frac{\alpha x^2}{2} \right]_0^L = Mg \left( \mu_0 L - \frac{\alpha L^2}{2} \right).$$

Given that $\mu_k(L) = 0$, we have $\mu_0 - \alpha L = 0$, leading to $\alpha = \frac{\mu_0}{L}$. Substituting this back gives:

$$W = Mg \left( \mu_0 L - \frac{\mu_0 L}{2} \right) = \frac{1}{2} \mu_0 Mg L.$$

Thus, $\eta = \frac{1}{2}$. Other options fail because they do not account for the correct integration of the varying friction force.