Question

A particle is projected along a horizontal field whose coefficient of friction varies as $\mu=A / r^{2}$, where $r$ is the distance from the origin in metres and $A$ is a positive constant. The initial distance of the particle is $1\, m$ from the origin and its velocity is radially outwards. The minimum initial velocity at this point so the particle never stops is

• A. $\infty$
• B. $2 \sqrt{g A}$
• C. $\sqrt{2 g A}$
• D. $4 \sqrt{g A}$

Answer 1

Answer: (C)

Work done against friction must equal to the initial kinetic energy.
$\frac{1}{2} m v^{2}=\int\limits_{1}^{\infty} \mu m g d x$
$\Rightarrow \frac{v^{2}}{2}=A g \int\limits_{1}^{\infty} \frac{1}{x^{2}} d x$
$\frac{v^{2}}{2}=A g\left[-\frac{1}{x}\right]_{1}^{\infty}$
$v^{2}=2 g A$
$\Rightarrow v=\sqrt{2 g A}$