1. Tardigrade
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  3. Physics
  4. 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

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Q. 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

Laws of Motion

Solution:

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