A particle is projected along a horizontal field whose coefficient of friction varies as \(mu = \frac{A}{r^2}\), where r is the distance from the origin in meters 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 :
Answer & explanation
Correct answer: option 2
Work done against friction is equivalent to the kinetic energy of the body :
\(\frac{1}{2}mv^2 = \int_1^{\infty} \mu .m. g .dr\)
\(\frac{v^2}{2} = \int_1^{\infty} \frac{A}{r^2} g dx\)
= \(-Ag\frac{1}{r}]_1^{\infty}\)
\(\Rightarrow v^2 = 2 Ag\)
\(\Rightarrow v = \sqrt{2Ag}\)