Two blocks `A` and `B` of masses `2kg` and `4kg` are placed one over the other as shown in figure. `A`time varying horizontal force `F =2t` is applied on the upper block as shown in figure. Here `t` is in second and `F` is in newton. Drow a graph showing acceleration of `A` and `B` on y-axis and time on x-axis.Coefficient of friction between `A` and `B` is `mu =(1)/(2)` and the horizontal surface over which `B` is placed smooth. `(g =10m//s^(2))`

Explanation

Limiting friction between `A` and `B` is
`f_(1) =mum_(A)g =(1)/(2) (2) (10) =10N`
Block `B` moves due to friction only. Therefore, maximum acceleration of `B` can be
`a_(max) =(f_(L))/(m_(B)) = (10)/(4) =2.5m//s^(2)`
Thus both the blocks move together with same acceleration becomes `25m//s^(2)`, after that acceleration of `B` will become constant while that of `A` will go on increasing. To find the time when the acceleration of both the blocks becomes `25m//s^(2)` (or when slipping will start between `A` and `B`) we will write

`2.5 =(F)/((m_(A)+m_(B)) = (2t)/(6)`
`:. t =7.5_(s)` Hence for `t le7.5_(s)`
`a_(A) =a_(B) = (F)/(m_(A)+m_(B)) = (2t)/(6) =(t)/(3)`
Thus, `a_(A)` versus `t` graph is a straight passing through origin of slop `(1)/(3)`
For, `tge7.5_(s)`
`a_(B) =2.5m//s^(2) = "constant"`
and `a_(A) =(F-f_(L))/(m_(A)`
or `a_(A) =(2t-10)/(2)` or `a_(A) =t-5`

Thus `a-(A)` versus `t` graph is a straight line of slop `1` and intercept`-5`. While `a_(B)` versus `t` graph is a strainght parallel to t-axis. The corresponding graph is as shown in above figure.