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

For the given situation shown in figure , choose the correct option `(

1.	For the given situation shown in figure , choose the correct option `(g = 10 m//s^(-2))` 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» Correct Answer - B::C `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.

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