Static and Kinetic Frictional Forces

AI Thread Summary
The discussion focuses on calculating the kinetic frictional force acting on a 6.00 kg box in different scenarios: stationary, accelerating upward, and accelerating downward in an elevator. The coefficient of kinetic friction is given as 0.360, which is used to determine the frictional force by dividing the actual friction force by the normal force. Participants clarify that the weight of the box is calculated as 58.8 N using the formula weight = mass × gravity. Confusion arises regarding how to incorporate the coefficient of kinetic friction and the effects of the elevator's acceleration on the net force acting on the box. The conversation emphasizes the importance of understanding the relationship between weight, normal force, and friction in dynamic situations.
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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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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?
 
Still Confused

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?
 
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