Block A weight `4N` and block B weight `8N` The coefficient of kinetic friction is `0.25` for all surface. Find F to slide B at constant speed when

A is held at rest

Explanation

a. If block `A and B` move together , the friction between `A and B` is static and friction between `B` and ground will be of kinetic nature.
`FBD` of `A +B` For `A and B` both move with constant velocity , the net force acting on the system of `A + B` should be zero
`F = f_(r) = mu N = mu(m_(A) + m_(B)) g`

b . If `A` is held at rest , The friction force on top and bottom surface of block `B` will be kinetic

Friction at the top surface of block `B`
`f_(1) = mu N_(1) = mu m _(A)g`
Friction at the bottom surface of block `B`
`f_(2) = mu N_(2) = mu (m _(A) + m_(B))g`
If block `B` moves with constant speed .
`F = f_(1) + f_(2) = mu m _(A)g + mu(m _(A) + m_(B))g`
`F = mu (2m_(A) + m_(B))A`
c. Now `A and B` are connected by light string .If force block `B` block `B` will move toward left block `A and B` are connected by string block `A` move toward right .The direction of friction force be opposite to relation motion and is shown in fig

From `FRD` of `A : T = f_(1) = mu N_(1) = mu m_(A)g`
From `FBD` of `B : F = f_(1) - f_(2) + T`
` = 2f_(1) + f_(2)`
`= 2 mu m_(A)g + mu (m_(A) + m_(B))g`
`rArr F = mu (3m_(A) + m_(B))A`