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

$(-\frac{k}{2r})$

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