Question

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

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

Answer 1

The correct option is A - $\frac{k}{2r}$

$\frac{m{v}^2}{r}$ = $\frac{k}{r^2}$ $\Rightarrow$ m $v^2$ = $\frac{k}{r}$ $\therefore$ K.E = $\frac{1}{2}m{v}^2$ = $\frac{k}{2r}$
P.E = $\int$F dr = $\int$ $\frac{k}{r^2}$dr = - $\frac{k}{r}$
$\therefore$ Total energy = K.E + P.E = $\frac{k}{2r}$ - $\frac{k}{r}$ = - $\frac{k}{2r}$