A force `F"` is applied on a block of mass `5 kg` as shown in figure. Co-efficient of friction between the wall and clock is `(sqrt(3))/(2). (g = 10m//s^(2))`

If `'f_(1)"` and `'f_(2)'` are magnitude of frictional force on the block when `F = 40 N` and `F = 140 N` repesctively. Then choose correct options

Options

A. `f_(1) + f_(2) = 50 N`
B. `(f_(1))/(f_(2)) = (3)/(2)`
C. `f_(2) - f_(1) = 90 N`
D. `(f_(1))/(f_(2)) = (3)/(5)`

Answer

A, B

Explanation

When `F = 40N, f = f_(1) = 30N`
`f_(L) = (sqrt(3))/(2) xx 70sqrt(3) = 105N`