Kinematics Question 742

Question: If a particle of mass $ m $ is moving in a horizontal circle of radius $ r $ with a centripetal force $ (-k/r^{2}) $ , the total energy it’s [EAMCET (Med.) 1995; AMU (Engg.) 2001]

Options:

A) $ -\frac{k}{2r} $

B) $ -\frac{k}{r} $

C) $ -\frac{2k}{r} $

D) $ -\frac{4k}{r} $

Show Answer

Answer:

Correct Answer: A

Solution:

$ \frac{mv^{2}}{r}=\frac{k}{r^{2}} $

therefore $ mv^{2}=\frac{k}{r} $

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

P.E.$ =\int{Fdr} $

$ =\int{{}}\frac{k}{r^{2}}dr=-\frac{k}{r} $

Total energy = K.E. + P.E. $ =\frac{k}{2r}-\frac{k}{r}=-\frac{k}{2r} $



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