SATHEE: Laws Of Motion Question 131

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) $2 F / 3$

C) $3 F / 5$

D) $5 F / 6$

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 $∴ T = [\frac{M + (M / 4)}{M + (M / 2)}] F = \frac{5 F}{6}$

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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:

Answer:

Solution:

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