The system starts from rest and `A` attains a velocity of `5m//s` after it has moved `5m` toward right. Assuming the arrangement to be frictionless everywhere and pulley and string to be light, the value of the constant force `F` applied on `A` is:

Options

A. `50N`
B. `75N`
C. `100N`
D. `96N`

Answer

B

Explanation

`a=(v^(2))/(2_(s))=(25)/(10)=2.5m//s^(2)`
For `6kg` : `-F-2T=6a`
For `2kg` : `-2g=2(2a)`
From (1) and (2) `F=75N` .