A small body of mass $m$ is projected with initial velocity $v_0$ from point $O$ on a rough horizont

Question

A small body of mass $m$ is projected with initial velocity $v_0$ from point $O$ on a rough horizontal plane. The coefficient of friction between the body and the plane increases linearly with distance $x$ from point $O$ as $\mu = \alpha x$, where $\alpha$ is a positive constant. What is the maximum instantaneous power developed by the friction force during the motion?

Professor PC Thomas Classes · Answer

The magnitude of the friction force at distance $x$ is $F_f = \mu mg = \alpha mg x$. By the work-energy theorem, $\frac{1}{2} m v^2 - \frac{1}{2} m v_0^2 = -\int_0^x F_f dx = -\frac{1}{2} \alpha mg x^2$. Solving for velocity yields $v(x) = \sqrt{v_0^2 - \alpha g x^2}$. The magnitude of instantaneous power is $P(x) = F_f v = \alpha mg x \sqrt{v_0^2 - \alpha g x^2}$. To maximize $P(x)$, we maximize $y = x^2(v_0^2 - \alpha g x^2)$. Setting $dy/d(x^2) = v_0^2 - 2\alpha g x^2 = 0$ gives $x = \frac{v_0}{\sqrt{2\alpha g}}$. At this position, $v = \frac{v_0}{\sqrt{2}}$, so $P_{max} = \alpha mg \left(\frac{v_0}{\sqrt{2\alpha g}}\right) \left(\frac{v_0}{\sqrt{2}}\right) = \frac{1}{2} m v_0^2 \sqrt{\alpha g}$.