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Coefficient of force

  1. Nov 18, 2007 #1
    1. The problem statement, all variables and given/known data
    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?


    2. Relevant equations

    coefficient=Ff/Fn
    Fn=mg

    3. 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?
     
  2. jcsd
  3. Nov 18, 2007 #2
    That is correct, but make sure to use two significant digits.
     
  4. Nov 18, 2007 #3
    Okay Thanks!
     
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