For the situation shown in figure ,mark the correct options.

Options

A. At `t = 3 s`,pseudo force on `4 kg` block applied from `2 kg` is`4 N` in forward direction
B. At `t = 3 s`,pseudo force on `2 kg` block applied from `4 kg` is `2 N` in forward direction
C. Pseudo force does not make an equal pairs
D. Pseudo force also makes a pair of equal and opposite forces

Answer

B, C

Explanation

Maximum force of friction between `2 kg and 4 kg`
`= 0.4 xx 2 xx 10 = 8 N`
`2 kg` move due to friction. Therefore its maximum acceleration may be
`a_(max) = (8)/(2) = 4 m//s^(2)`
slip will start when their combined acceleration becomes `4 m//s^(2)`
`:. a = (F)/(m)` or `4 = (2t)/(6)` or `t = 12 s`
`A t = 3 s`
`a_(2) = a_(4) = (F)/(m) = (2t)/(6) = (2 xx 3)/(6)`
`= 1 m//s^(2)`
Both `a_(2) and a_(4)` are towards right. Therefore pseudo force `F_(1)` (on `2 kg` from `4 kg`) and `F_(2)` (on `4 kg` from `2 kg)` are towards left
`F_(1) = (2) (1) = 2 N`
`F_(2) = (4) (1) = 4 N`
From here we can see that `F_(1) and F_(2)` do not make a pair of equal and opposite force.