Question

A particle is moving in a circular path of radius a under the action of an attractive potential energy, $U=-\frac{k}{2r^2}$. Its total energy is

• A. $\frac{k}{4a^2}$
• B. $\frac{k}{2a^2}$
• C. zero
• D. $-\frac{3}{2}\frac{k}{a^2}$

Answer 1

The correct option is C zero

We know that,

The force is

$F=\frac{m{v}^2}{r}=\frac{k}{r^3}$

$m{v}^2=\frac{k}{r^2}$

Kinetic energy is

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

$K.E=\frac{k}{2r^2}$

Total energy is

$E_{total}=P.E+K.E$

$E_{total}=-\frac{k}{2r^2}+\frac{k}{2r^2}$

$E_{total}=0$

Hence, the total energy is $0$