QJEEM18Ph85b9: JEE Main 2018 Physics - Class 11 JEEM18Ph85b9 Solution: A particle is moving in a circular path of radius under the…

We know, Total energy = Kinetic energy + Potential energy

Potential energy given as $U = - {k \over {2{r^2}}}$

We need to find Kinetic Energy.

As Force acting on the particle (F) = $- {{dU} \over {dr}}$

$\Rightarrow F = - {d \over {dr}}\left( {{{ - k} \over {2{r^2}}}} \right)$

$= {k \over 2} \times \left( { - 2} \right) \times {r^{ - 3}}$

$= - {k \over {{r^3}}}$

Because of this force particle is having circular motion so it will provide possible centripetal force.

$\left| F \right| = {{m{v^2}} \over r}$

$\Rightarrow {{m{v^2}} \over r} = {k \over {{r^3}}}$

$\Rightarrow$ $m{v^2} = {k \over {{r^2}}}$

We know kinetic energy of particle, K = ${1 \over 2}m{v^2}$ = ${k \over {2{r^2}}}$

As Total energy = Kinetic energy + Potential energy

So Total energy = ${k \over {2{r^2}}}$ $- {k \over {2{r^2}}}$ = 0