InterviewSolution
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InterviewSolution
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For the situation shown in figure ,mark the correct options. A. At `t = 3 s`,pseudo force on `4 kg` block applied from `2 kg` is`4 N` in forward directionB. At `t = 3 s`,pseudo force on `2 kg` block applied from `4 kg` is `2 N` in forward directionC. Pseudo force does not make an equal pairsD. Pseudo force also makes a pair of equal and opposite forces |
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Answer» Correct Answer - B::C 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. |
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