Static and Kinetic Frictional Forces

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SUMMARY

The discussion focuses on calculating the kinetic frictional force acting on a 6.00 kg box sliding across a horizontal floor in an elevator under various conditions. The coefficient of kinetic friction is 0.360. When the elevator is stationary, the kinetic frictional force is determined using the formula: friction force = coefficient of kinetic friction × normal force. The normal force varies based on the elevator's acceleration, affecting the net force on the box during upward and downward acceleration scenarios.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concept of normal force
  • Knowledge of kinetic friction and its coefficient
  • Basic algebra for force calculations
NEXT STEPS
  • Calculate the normal force in different acceleration scenarios
  • Explore the relationship between weight and normal force
  • Learn about static friction and its applications
  • Study the effects of varying coefficients of friction on motion
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Students studying physics, particularly those focusing on mechanics and forces, as well as educators looking for practical examples of friction in real-world scenarios.

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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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