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