Static and Kinetic Frictional Forces

A 6.00 kg box is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the box and the floor is 0.360. Determine the kinetic frictional force that acts on the box when the elevator is (a) stationary, (b) accelerating upward with an acceleration whose magnitude is 1.20 m/s2, and (c) accelerating downward with an acceleration whose magnitude is 1.20 m/s2.
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 Recognitions: Gold Member Science Advisor Staff Emeritus Why did you title this "Static and Kinetic Friction Forces" when there are no static friction forces involved? The "coefficient of kinetic friction force" is defined as "actual friction force divided by normal force". On a flat floor, the "normal force" is just the net vertical force. What is the weight of a 6.00 kg box (the downward force due to gravity)? If the acceleration is 1.2 m/s2 upward, what additional downward for is there? (F= ma)? What is the net force on the box? If the acceleration is 1.2m/s2 downward, what is the "upward" force? What is the net force on the box?
 I'm super confused. I got lost after you said find weight. Isn't weight mg? So that would be 6*9.8=58.8N, but I don't know where to go from there. Where does the .360 come in?