A rope of length $L$ and mass $M$ is being pulled on a rough horizontal floor by a constant horizontal force $F$ = $Mg$ . 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 midpoint of the rope is
A$\frac{{Mg}}{4}$
B$\frac{{2Mg}}{5}$
C$\frac{{Mg}}{8}$
D$\frac{{Mg}}{2}$
Diffcult
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D$\frac{{Mg}}{2}$
d $a=\frac{F-\mu N}{M}=\frac{M g-0.5 M g}{M}=g / 2$
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