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)=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)`

`F=10N`
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 become 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 `tle7.5s`
`a_(A)=a_(B)=(F)/(m_(A)+m_(B))=(2t)/(6)=(t)/(3)`