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} $
BETA
