Calculating force for constant speed: Blocks with friction and a pulley

[courtrigrad]

Block A is on top of Block B. They are connected by a light, flexible cord passing through a fixed, frictionless pulley. Block A weighs 3.60 N and block B weighs 5.40 N. The coefficient of kinetic friction between all surfaces is 0.25. Find the magnitude of the force F necessary to drag Block B to the left at a constant speed.
Ok so the key here is constant speed. So that means F = 0. So all forces must be equal and opposite one another in the same direction. Since w_{total} = 9.0 N does that mean that the force required is f_{k} = \mu_{k}N = 0.25(9.0 N) = 2.5 N?
Thanks

 

[Astronuc]

You have part of the answer.

Block A is on top of block B, and they are connected by a pulley. If B moves left, A moves right, and the friction force of A's motion resists B's motion.

Block B is on a surface, also with \mu = 0.25, but the downward force arises from the masses of A and B.

Now at constant velocity, there is not acceleration, so the net force F = 0, so F = Friction force of block A on B + Friction force of block B on the surface underneath.

 

[courtrigrad]

so it would be 0.25(3.6 N) + 0.25(5.40 N) = 2.25 N. I forgot the fact that when you pull on block A, block B goes in the opposite direction.

Thanks