Work Energy And Power Question 138
Question: A mass m starting from A reaches B of a frictionless track. On reaching B, it pushes the track with a force equal to x times its weight, then the applicable relation is
Options:
A) $ h=\frac{(x+5)}{2}r $
B) $ h=\frac{x}{2}r $
C) $ h=r $
D) $ h=( \frac{x+1}{2} )r $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] KE of Blocks at $ B=PE $ at A-PE at B $ \frac{1}{2}mv^{2}=mgh-mg2r=mg(h-2r) $
$ v^{2}=2g(h-2r) $ (i)
Also, $ \frac{mv^{2}}{r}=xmg+mg $
Or $ v^{2}=(x+1)rg $ (ii) Equating Eqs. (i) and (ii), we get $ 2g(h-2r)=(x+1)gr $ Or $ 2gh=(x+1)gr+4gr=(x+5)gr $
$ h=( \frac{x+5}{2} )r $
