Calculating Coefficient of Force for 2.6 kg Box on Concrete Floor

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A force of 22.0 N is needed to initiate movement of a 2.6 kg box on a concrete floor, resulting in an acceleration of 0.50 m/s². The coefficient of kinetic friction is calculated using the formula coefficient = Ff/Fn, where Fn is the normal force (Fn = mg). The frictional force (Ff) is determined by first calculating the net force (F = ma), yielding 1.3 N, and then subtracting this from the applied force, resulting in a frictional force of 20.7 N. The coefficient of kinetic friction is then found to be approximately 0.812, which should be reported with two significant digits. The calculations confirm the accuracy of the coefficient.
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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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