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} $
