Two blocks `A(m_(A) = 5 kg)` and `B(m_(B)= 15 kg)` are placed as shown in Fig A variable force `F = 200` starts actsing from time `t = 0` on lower bolck B just large anough to Determine the force `F` to make block B sliding out from between the blocks A and the ground at this instant , plot a graph between acceleration of both the blocks and time

Explanation

If upper block start sliding w.r.t lower block the friction between the surface os A and B should be limiting
Let `f_(A)` and `f_(B)` are friction between A and B respectively maximum friction between A and B
= Limiting friction between A and B
`rArr (f_(A))_(max) = (f_(A))_(lim)`
`= mu_(s)N_(A)= mu_(s)m_(A) g`
` = 03 xx 5 xx 10 = 15 N`
and kinetic friction between A and B

`(f_(A))_("lim") = mu_(s)N_(A) g = 0.1 xx 5 xx 10 = 5 N`
Similary for block B and ground
`(f_(B))_("max") = mu_(k) N_(B) = mu_(s) (N_(A) +m_(B)g) = mu_(s)(m_(A)g + m_(B)g)`
or `(f_(B))_("max") = mu_(s) (m_(A) + m_(B))g = 0.5(5 + 15) xx 10= 100 N`
`(f_(B))_("lim") = mu_(s) (m_(A) + m_(B))g = 0.4( 5 + 15 ) xx 10 = 80 N`
If `F gt 100 N` (or after `5 s)` the block B will start sliding with respect to ground and friction appeart the surface of A and B
Upto the time when friction between the blocks A and B reaches ti its limiting value , both the blocks move with different acceleration
Acceleration of block A when slipping between
`((f_(A))_(max))/(m_(A)) = (15)/(5) = 3 ms^(-2)`
As this time, blocks B is sliding on the ground friction between block B and ground is kinetic nature

From FBD equation of motion of blocks B
`f - [(f_(A))_("lim") + (f_(B))_("lim")]= m_(B) a`
` F - [15 + 80] = 15 xx 3`
`rArr F = 140 N`
It occursat time `t = (140)/(20)= 7 s` (time `t gt 7 s)`
For force range`100 N lt F le 140 N`
or` 100 N lt 20t le 140 N`
i. e. For time range` 5 s lt r le 7 s` both blocks move with cammon acceleration
` a_(A)= a_(B) = (F- (f_(B))_("lim"))/((m _(A) + m_(B)) = (20 t - 80)/(20) = (t -4 )ms^(-2)`