If a particle of mass m is moving in a horizo | vedclass

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(A) The centripetal force is given by $F = \frac{mv^2}{r} = \frac{k}{r^2}$.From this,we get $mv^2 = \frac{k}{r}$.The kinetic energy $(K.E.)$ is given

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

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(A) The centripetal force is given by $F = \frac{mv^2}{r} = \frac{k}{r^2}$.From this,we get $mv^2 = \frac{k}{r}$.The kinetic energy $(K.E.)$ is given by $K.E. = \frac{1}{2}mv^2 = \frac{k}{2r}$.The potential energy $(P.E.)$ is found by integrating the force: $P.E. = -\int F dr = -\int \frac{k}{r^2} dr = -\frac{k}{r}$.The total energy $(E)$ is the sum of kinetic and potential energy: $E = K.E. + P.E. = \frac{k}{2r} - \frac{k}{r} = -\frac{k}{2r}$.

If a particle of mass $m$ is moving in a horizontal circle of radius $r$ with a centripetal force $(-k/r^2)$,the total energy is

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