Work Energy And Power Question 63

Question: A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to $ -K/r^{2} $ , where K is a constant. The total energy of the particle is [IIT 1977]

Options:

A) $ \frac{K}{2r} $

B) $ -\frac{K}{2r} $

C) $ -\frac{K}{r} $

D) $ \frac{K}{r} $

Show Answer

Answer:

Correct Answer: B

Solution:

Here $ \frac{mv^{2}}{r}=\frac{K}{r^{2}} $

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

$ U=-\int _{\infty }^{r}{F.dr}=-\int _{\infty }^{r}{( -\frac{K}{r^{2}} )}dr=-\frac{K}{r} $

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



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