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A particle is moving in a circular path of radius a under the action of an attractive potential U = -k/(2r²). Its total energy is:

1. -k/(4a²)
2. k/(2a²)
3. zero
4. -3k/(2a²)

Correct answer: zero

Solution

F = -dU/dr = -k/r^3 (attractive). Circular motion: mv^2/r = k/r^3, so KE = (1/2)mv^2 = k/(2a^2). Total E = KE + U = k/(2a^2) - k/(2a^2) = 0.

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