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

• A. F
• B. 2 F/3
• C. 3 F/5
• D. 5 F/6

Answer 1

Answer: (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 $m=M/2$
$\therefore T=\left[\frac{M+(M/4)}{M+(M/2)}\right]F=\frac{5F}{6}$