For the given situation shown in figure , choose the correct option `(g = 10 m//s^(-2))`

Options

A. At `t = 1 s`, force of friction between `2 kg` and `4 kg` is `2 N`
B. At `t = 1 s`, force of friction between `2 kg` and `4 kg` is zero`
C. At `t = 4 s`, force of friction between `4 kg` and ground is `8 N`
D. At `t = 15 s`, acceleration of `2 kg 1ms^(-2)`

Answer

B, C

Explanation

`f_(1) rarr` force of friction between`2 kg` and` 4 kg`
`f_(2) rarr` force of friction between 4 kg and ground
`(fs_(1))_(max) = 0.4 xx 2 xx 10 = 8 N`
`Fk_(1) = 0.2 xx 2 xx 10 = 4 N`
`(fs_(2))_(max) = 0.6 xx 6 xx 10 = 36 N`
`Fk_(2) = 0.4 xx 6 xx 10 = 24 N`
At `t = 1 s`, `F = 2 N lt 36 N`, therefore system remains stationary and force of friction between `2 kg` and `4 kg` is zero
At `t = 4 s`, `F = 8 N lt 36 N`, therefore system is again stationary and force of friction between `4 kg` from ground is `8 N`
At `t = 15 s`, `F = 30 N lt 36 N`, therefore system remains stationary and force of friction between `2 kg` and `4 kg` is zero
At `t = 1 s`, `F = 2 N lt 36 N`, and system is stationary.