Q. If a particle of mass $m$ is moving in a horizontal circle of radius $r$ with a centripetal force $\left( - k/r^{2} \right),\ $the total energy is
A) $- \frac{k}{2r}$
B) $- \frac{k}{r}$
C) $- \frac{2k}{r}$
D) $- \frac{4k}{r}$
Explanation
(a) $\frac{mv^{2}}{r} = \frac{k}{r^{2}}\ \Rightarrow mv^{2} = \frac{k}{r}\ \therefore K.E.\ = \frac{1}{2}\ mv^{2} = \frac{k}{2r}$
P.E. $= \ \int_{}^{}{F\ dr} = \int_{}^{}\frac{k}{r^{2}}\ dr = - \frac{k}{r}$
$\therefore$ Total energy $= K.E.\ + P.E.\ = \frac{k}{2r} - \frac{k}{r} = - \frac{k}{2r}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.