Calculating Coefficient of Force for 2.6 kg Box on Concrete Floor

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SUMMARY

The discussion focuses on calculating the coefficient of kinetic friction for a 2.6 kg box on a concrete floor, requiring a force of 22.0 N to initiate movement. Using Newton's second law, the acceleration of the box is determined to be 0.50 m/s², leading to a frictional force (Ff) of 20.7 N after accounting for the net force. The coefficient of kinetic friction is calculated as 0.812, confirming the correct application of the formula coefficient = Ff/Fn, where Fn is the normal force derived from the box's weight (Fn = mg).

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  • Understanding of Newton's second law (F=ma)
  • Knowledge of frictional force calculations
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  • Basic proficiency in significant figures in calculations
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Homework Statement


A force of 22.0 N is required to start a 2.6 kg box moving across a horizontal concrete floor.
1)If the 22.0 N force continues, the box accelerates at 0.50 m/s2. What is the coefficient of kinetic friction?


Homework Equations



coefficient=Ff/Fn
Fn=mg

The Attempt at a Solution



To find Ff, I used F=ma and got (2.6)*(0.50) and got 1.3. I subtracted the 1.3 from the 22.0 N and got 20.7 then
i used Ff/Fn which was 20.7/25.48 and got .812 is that correct?
 
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That is correct, but make sure to use two significant digits.
 
Okay Thanks!
 

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