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

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Question

(a) $\frac{{m{v^2}}}{r} = \frac{k}{{{r^2}}}$ $&rArr;$ $m{v^2} = \frac{k}{r}$

$	herefore K.E.$ = $\frac{1}{2}m{v^2} = \frac{k}{{2r}}$

$P.E. = \i

Vedclass · Accepted Answer

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

Vedclass · Answer

(a) $\frac{{m{v^2}}}{r} = \frac{k}{{{r^2}}}$ $&rArr;$ $m{v^2} = \frac{k}{r}$

$	herefore K.E.$ = $\frac{1}{2}m{v^2} = \frac{k}{{2r}}$

$P.E. = \int {F\,dr} $ $ = \int {} \frac{k}{{{r^2}}}dr = - \frac{k}{r}$

Total energy $= K.E. + P.E.$ $ = \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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