Question

A particle is moving in a circular path of radius a under the action of an attractive potential U=-(k)/(2r^(2)). Its total energy is :

Answer 1

Answer:

zero

Explanation:

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

hope it helps you mark as brainliest please