Question

If a particle of mass m is moving in a horizontal circle of radius $ r $ with a centripetal force $ \left(-\frac{K}{r^{2}}\right) $ , total energy is

• A. $ -\frac{K}{2r} $
• B. $ -\frac{K}{r} $
• C. $ -\frac{2K}{r} $
• D. $ -\frac{4K}{r} $

Answer 1

Answer: (A)

Given, centripetal force, $F = -\frac{K}{r^2}$
Also, $\frac{K}{r^2} = \frac{mv^2}{r} $ or $mv^2 = \frac{K}{r}$
$\therefore $ Kinetic energy, $K.E. = \frac{1}{2} mv^2 = \frac{K}{2r}$
For central forces, $P.E. = - 2 K.E. = - \frac{K}{r}$
Total energy, $T.E. = P.E. + K.E.$
$= \frac{-K}{r} + \frac{K}{2r} = - \frac{K}{2r}$