Find the contact force between the 3kg and 3kg block as shown in figure.

Explanation

Considering both blocks as a system to find the common acceleration
`F_("net") = F_(1) - F_(2) = 100 -25 = 75N`
common acceleration
`a = ( F_("net"))/( 5kg) = ( 75)/( 5) = 15 m//s^(2)`

To find the contact force between A & B we draw F.B.D. of 2 kg block from `( sum F_("net"))_(x) = ma_(x)`

`rArr N-25 = ( 2) ( 15)`
`rArr N = 55N`
Case (ii) `:` Three body system `:`

Free body diagrams `:`
`rArr a = ( F )/( m_(1) +m_(2) + m_(3))`
and `f_(1) = ((m_(2) + m_(3))f)/( (m_(1) + m_(2)+m_(3)))`

`f_(2) =( m_(3) F)/((m_(1) + m_(2) +m_(3)))`
`f_(1) = ` contact force between masses `m_(1) ` and `m_(2)`
`f_(2) = `contact force between masses `m_(2)` and `m_(3)`
Remember `:` Contact forces will be different if force F will be applied on mass C