Work Energy And Power Question 305
Question: A simple pendulum with a bob of mass ’m’ oscillates from A to C and back to A such that PB is H. If the acceleration due to gravity is ‘g’, then the velocity of the bob as it passes through B is
Options:
A) zero
B) $ 2gH $
C) $ mgH $
D) $ \sqrt{2gH} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] At B, the velocity is maximum.
Taking vertical downward motion of body from A to B, we have, $ u=0,a=g,s=H $
and $ v=? $
$ v^{2}=u^{2}+2as; $
or $ v^{2}=0+2gH $ or $ v=\sqrt{2gH} $
BETA
