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2 months, 3 weeks ago

# A particle is moving in a circular path of radius a under the action of an attractive potential U= Â . Its total energy is :

IMG_20190128_102517.jpg

A particle is moving in a circular path of radius a under the action of an attractive potential U=  . Its total energy is :

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A Avinash Dongre
Answered 2 months, 3 weeks ago

@Shilpa

$U=-\frac{K}{2r^{2}}$

$\therefore F=\frac{-dU}{dr}=\frac{-K}{r^{3}}$

$\Rightarrow \frac{K}{r^{3}}=\frac{mv^{2}}{r} or$  $mv^{2}=\frac{K}{r^{2}}$

$K.E.=\frac{1}{2}mv^{2}=\frac{K}{2r^{2}}$

Total energy = K.E.+P.E.=0

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