Question

A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to $\frac{-K}{r^2}$ , where K is a constant. The total energy of the particle is

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

Answer 1

The correct option is B

$-\frac{K}{2r}$

$Here\frac{m{v}^2}{r}=\frac{K}{r^2}\therefore K.E=\frac{1}{2}m{v}^2=\frac{K}{2r}$

$U=-\underset{\infty }{\overset{r}{\int }}F.dr=-{\int }_{\infty }^r(-\frac{K}{r^2})dr=-\frac{k}{r}$

Total energy $E=K.E+P.E=\frac{K}{2r}-\frac{K}{r}=-\frac{K}{2r}$