A rope of length `L` and mass `M` is being pulled on a rought horizontal floor by a contact horizontal force `F = m g`. The force is acting at one end of the rope in the same direction as the length of the rop. The coefficient of kinetic friction between rope and floor is `1//2`. Then the tension at the midpoint of the rop is

Options

A. `(M g)/(4)`
B. `(2 M g)/(5)`
C. `(M g)/(8)`
D. `(M g)/(2)`

Answer

D

Explanation

`a = (F - f)/(M)`

`T - mu ((M)/(2)) g = (M)/(2)`
`a = ((M)/(2)) (g)/(2)`
Putting `mu = (1)/(2)`, we get
`T = (M g)/(2)`