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May the Force be 81.2 N

  1. Jan 29, 2014 #1
    1. The problem statement, all variables and given/known data

    A 15 kg crate rests on the floor. The coefficient of friction between the crate and the floor is 0.28. How much horizontal force is needed to accelerate the crate at 3.0 m/s^2?


    2. Relevant equations

    Fnet=ma

    3. The attempt at a solution
    m=15 kg
    F-f=ma
    F-[itex]M[/itex]N=ma
    F-[itex]M[/itex]mg=ma
    F=ma+[itex]M[/itex]mg
    F=15*3+.38*15*9.8
    F=81.16 N.
    For a final rounded answer of: F=81.2 N.

    This answer seems correct to me, could someone please confirm?

    Thanks! :smile:
     
  2. jcsd
  3. Jan 29, 2014 #2

    SteamKing

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    I don't get your result, even with μ = 0.28. Try your calculator again.
     
  4. Jan 29, 2014 #3
    Oops, I thought I corrected that, sorry! Weird, I somehow mixed up the digits. Does 86.16, rounded to 86.2, sound better?

    Thanks!
     
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