Work Energy And Power Question 329
The particle is released from a height h. At a certain height, its KE is two times its potential energy. Height and speed of the particle at that instant are
Options:
A) $ \frac{h}{3},\frac{\sqrt{2gh}}{3} $
B) $ \frac{h}{3},\frac{\sqrt{gh}}{3} $
C) $ \frac{2h}{3},\frac{\sqrt{2gh}}{3} $
D) $ \frac{h}{3},\sqrt{2gh} $
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Answer:
Correct Answer: B
Solution:
[b] Total mechanical energy = mgh + ½mv²
As, $ \frac{KE}{PE}=\frac{2}{1} $
$ \frac{KE}{PE}=\frac{2}{1} $ mgh and $ PE=\frac{2}{3}mgh $
Height from the ground at this instant,
$ h’=\frac{h}{3}, $ and speed of particle at this instant,
$ v=\sqrt{2g(h-h’)}=\sqrt{2g( \frac{2h}{3} )}=2\sqrt{\frac{gh}{3}} $
BETA
