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. $2\sqrt{gA}$
• C. $\sqrt{2gA}$
• D. $4\sqrt{gA}$

Answer 1

Answer: (C)

$\sqrt{2gA}$

Work done against friction is equivalent to the kinetic energy of the body

$\frac{1}{2}m{v}^2=\underset{1}{\overset{\infty }{\int }}\mu mgdr$

$\frac{v^2}{2}=\underset{1}{\overset{\infty }{\int }}\frac{A}{r^2}gdx$

$\frac{v^2}{2}=-Ag{{\left[\frac{1}{r}\right]}_1}^{\infty }$

$v^2=2Ag$

$v=\sqrt{2Ag}$