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

Answer 1

The correct option is A zero

$U=-\frac{K}{2r^2}=P.E$
$F=-\frac{du}{dr}=-(-\frac{K}{2}(-\frac{2}{r^3}))=-\frac{K}{r^3}$
$\frac{K}{r^3}=\frac{m{v}^2}{r}\Rightarrow m{v}^2=\frac{K}{r^2}$
$K.E.=\frac{1}{2}m{v}^2=\frac{K}{2r^2}$
$E=P.E.+K.E.=0$