Calculating Coefficient of Force for 2.6 kg Box on Concrete Floor

[mortho]

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?

 

[hotcommodity]

That is correct, but make sure to use two significant digits.

 

[mortho]

Okay Thanks!