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A block is lying on a 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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$\frac{2 F}{3}$

$\frac{3 F}{5}$

$\frac{5 F}{6}$

answer is D.

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Detailed Solution

Conceptual

Explanation: acceleration of the system= a= $\frac{A p p l i e d f o r c e}{t o t a l m a s s o f t h e s y s t e m} = \frac{F}{\frac{3 m}{2}} = \frac{2 F}{3 m}$

Tension in the rope at the middle of the rope is given by F- T= ma/4

so T = $(m + \frac{m}{4}) a = \frac{5 m}{4} \times \frac{2 F}{3 m} = \frac{5 F}{6}$

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