AP EAMCET · PHYSICS · Motion In Two Dimensions
A particle is moving along a horizontal circle of radius ' \(r\) ' under a centripetal force \(\frac{-c}{r^2}\) where ' \(c\) ' is a constant. Then, the total energy of the particle is
- A \(\frac{-c}{2 r^2}\)
- B \(\frac{c}{2 r}\)
- C \(\frac{-c}{2 r}\)
- D \(\frac{c}{2 r^2}\)
Answer & Solution
Correct Answer
(C) \(\frac{-c}{2 r}\)
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