Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Work Power Energy

Question:

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 :

Options:

\(2 \sqrt{gA}\)

\(\sqrt{2gA}\)

\(4 \sqrt{gA}\)

\(\infty\)

Correct Answer:

\(\sqrt{2gA}\)

Explanation:

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

\(\frac{1}{2}mv^2 = \int_1^{\infty} \mu .m. g .dr\)

\(\frac{v^2}{2} = \int_1^{\infty} \frac{A}{r^2} g dx\)

  = \(-Ag\frac{1}{r}]_1^{\infty}\)

\(\Rightarrow v^2 = 2 Ag\)

\(\Rightarrow v = \sqrt{2Ag}\)