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A rope of length L and mass M is being pulled on a rough horizontal floor by a constant horizontal force $F = M g .$ The force is acting at one end of the rope in the same direction as the length of the rope. The coefficient of kinetic friction between rope and floor is 1/2. Then the tension at the mid-point of the rope is

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$\frac{M g}{4}$

$\frac{2 M g}{5}$

$\frac{M g}{8}$

$\frac{M g}{2}$

answer is D.

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Detailed Solution

At mid point, $T - μ \frac{M}{2} g = \frac{M}{2} a - (1), M g - μ M g = M a - (2)$

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