Question

A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to (-k)/r^2 where k is a constant. what is the total energy of the particle?

Answer 1

Given, centripetal force = $\frac{mv^2}{r}$= $-\frac{k}{r^2}$

so, $mv^2=\frac{-k}{r}$

kinetic energy = $\frac{1}{2}mv^2=\frac{-k}{2r}$.........(1)

now, potential energy = $-\int\limits^r_{\infty}{F}\,dr$

= $-\int\limits^r_{\infty}{\frac{-k}{r^2}}\,dr$

= $k\left|\frac{r^{-1}}{-1}\right|^r_{\infty}$

=$\frac{-k}{r}$

now, total energy = kinetic energy+ potential energy
= $\frac{-k}{2r}+\frac{-k}{r}$

= $\frac{-3k}{2r}$

Answer 2

Answer:

-k/2r

Explanation:

negative energy means that particle is in bound state