Two blocks `A` and `B` of mass `2 kg` and `4 kg` are placed one over the others as shown in figure. A time vaying horizontal force `F=2t` is applied on the upper blocks as shown in figure. Here `t` is in second and `F` is in newton. Coefficient of friction between `A` and `B` is `mu=(1)/(2)` and the horizontal surface over which `B` is placed is smmoth. `(g=10 m//s^(2))`. If acceleration of blocks `A` as a function of time is given by `a_(A)=t//c` then find value of `c`. `(t le7.5s)`

Answer

3

Explanation

Limiting friction between A and B is
`f_(L) = mu m_(A) g ((1)/(2)) (2)(10) = 10 N`

Block B moves due to friction only . Therefore , maximum acceleration of B can be
`a_("max") = (f_(L))/(m_(B)) = (10)/(4) = 2.5 m//s^(2)`
Thus , both the blocks move together with same acceleration till the common acceleration becomes 2.5 `m//s^(2)` , after that acceleration of B will becomes constant while that of A will go on increasing . To find the time when the acceleration of both the blocks becomes `2.5 m//s^(2)` (or when slipping will start between A and B) we will write
`2.5 = (F)/((m_(A) + m_(B))) = (2t)/(6)`
Therefore t = 7.5 s
Hence , for `t le 7.5 s`
`a_(A) = a_(B) = (F)/(m_(A) + m_(B)) = (2t)/(6) = (t)/(3)`