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
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$a=\frac{F-\mu N}{M}=\frac{M g-0.5 M g}{M}=g / 2$
$\mathrm{T}-\mu \mathrm{Mg} / 2=\mathrm{Ma} / 2$
$\mathrm{T}-\mathrm{mg} / 4=\mathrm{Mg} / 4 \Rightarrow \mathrm{T}=\mathrm{Mg} / 2$
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