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  • A particle is moving in a circular path of radius a under the action of an attractive potential

A particle is moving in a circular path of radius a under the action of an attractive potential U =?K/2r^2 . what is total energy

Answers (2)

Total energy is zero 

Posted by

ABHOUMA S SUNIL

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its simple 

as we know 

F = dU/dr

F = d(-k/2r^2)/dr

F = k/r^3

as we know F = mv^2/r

mv^2/r = k/r^3

mv^2 = k/r^2

1/2 mv^2 = 1/2 k/r^2

K.E = 1/2 k/r^2

and as given U = -k/2r^2

total energy = K.E  + U

total energy = zero

 

Posted by

Rajat

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