Laws Of Motion Question 131
Question: A block is lying on the horizontal frictionless surface. One end of a uniform rope is fixed to the block which is pulled in the horizontal direction by applying a force F at the other end. If the mass of the rope is half the mass of the block, the tension in the middle of the rope will be
Options:
A) $ F $
B) $ 2F/3 $
C) $ 3F/5 $
D) $ 5F/6 $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] The acceleration of block-rope system is $ a=\frac{F}{(M+m)} $
Where M is the mass of block and m is the mass of rope.
So the tension in the middle of the rope will be $ T={M+(m/2)}a=\frac{M+(m/2)F}{M+m} $ Given that m=M/2 $ \therefore T=[ \frac{M+(M/4)}{M+(M/2)} ]F=\frac{5F}{6} $
BETA
