Question

A particle of mass $m$, located on a horizontal surface at a point $O$, acquire a horizontal velocity $u$. If the friction coefficient varies as $\mu =\alpha x$, where $\alpha$ is a constant and $x$ is the distance from the point $O$, find the value of $x$ for which the instantaneous power developed by the friction is maximum.

• A. $x=\frac{u}{2\alpha g}$
• B. $x=\frac{u}{\sqrt{\alpha g}}$
• C. $x=\frac{u}{\sqrt{2\alpha g}}$
• D. $x=\frac{u}{2\sqrt{\alpha g}}$

Answer 1

The correct option is C $x=\frac{u}{\sqrt{2\alpha g}}$

When friction coefficient is $\mu =\alpha x$, the friction force on the body when it is at a distance $x$ from the point $O$ is
$f=\mu N=\alpha xmg[\because N=mg]$

Retardation due to this force:
$a=-\frac{f}{m}=-\alpha gx$

Now, $\frac{vdv}{dx}=-\alpha gx\Rightarrow vdv=-\alpha gxdx$

$\Rightarrow {\int }_u^{v}vdv=-\alpha g{\int }_0^{x}xdx$

$\Rightarrow v^2=u^2-\alpha g{x}^2$

We know that, $P_{inst}=F\cdot v=-\alpha mgx\left(\sqrt{u^2-\alpha g{x}^2}\right)$

The power is maximumm when $\frac{dP}{dx}=0$.

$\Rightarrow \frac{dP}{dx}=-[\frac{\alpha mgx}{u^2-\alpha g{x}^2}{]}^{\frac{1}{2}}\times \alpha gx-\alpha mg[u^2-\alpha g{x}^2{]}^{\frac{1}{2}}=0$

$\Rightarrow x=\frac{u}{\sqrt{2\alpha g}}$