Question

A particle of mass $m$ is moving in circular path only under action of central force whose potential energy is $V=-mk{r}^5$, here $r$ is distance from centre and $k$ is constant. Then kinetic energy of particle is equal to

• A. $\frac{5}{2}mk{r}^5$
• B. $5mk{r}^5$
• C. $\frac{5}{2}mk{r}^{5/2}$
• D. $5mk{r}^{5/2}$

Answer 1

The correct option is A $\frac{5}{2}mk{r}^5$

$U-km{r}^5$
$F=-\frac{dU}{dr}=+5km{r}^4$
$F=F_c=\frac{m{v}^2}{r}$
$5km{r}^4=\frac{m{v}^2}{r}$
$\frac{m{v}^2}{2}=\frac{5}{2}km{r}^5$
K.E =$\frac{5}{2}km{r}^5$