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Forces AND Friction AND Motion O MY!

  1. Oct 19, 2009 #1
    1. The problem statement, all variables and given/known data

    A cart with a mass of 285 kg is pushed horizontally with a force of 1400 N. The acceleration of gravity is 9.8 m/s/s. The coefficient of friction (u) with the crate and the surface is 0.29. Calculate the acceleration of the crate.

    Sooo given:

    Fa: 1400 N
    m: 285 kg
    Fg: -9.8
    u: 0.29
    Ff: ?
    Fn: ?
    a: ?


    2. Relevant equations

    Fnet = ma
    Ff = u * Fn

    3. The attempt at a solution

    Ok....
    so if Fg is -9.8 then Fn is 9.8?
    and then Ff = -(0.29)(9.8) which is -2.842?
    so would Fnet = 1397.158 N?
    and then do I use F = ma to find a?
    so a = F/m which i got 4.9023
    but the answer in the back of my text book is 2.07028!
    help please!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 20, 2009 #2

    tiny-tim

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    Hi kza62! :wink:
    no no no no noooo :cry:

    -9.8 is the acceleration, isn't it? :smile:
     
  4. Oct 25, 2009 #3
    not really dude, its moving horizontally, its not falling
     
  5. Oct 26, 2009 #4

    tiny-tim

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    Yes, but I meant that when you wrote Fg = -9.8, I assumed that your F meant a force, so that would be -9.8m :wink:
     
  6. Oct 26, 2009 #5
    ooooo okokokok but how do u get Fn?? thats the only thing i need to plug in everything else :uhh:
     
  7. Oct 27, 2009 #6

    tiny-tim

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    (just got up :zzz: …)

    The ground is horizontal, so Fn (the normal force) is the same as Fg, and then Ffriction = µFn.

    You know that, don't you? :smile:
     
  8. Nov 1, 2009 #7
    so Ff = 0.29 x 9.8 = 2.842
    and then Fnet = 1400 - 2.842 = 1397.158
    a = 1397.158/285 = 4.902308... ????

    what did i do wrong now???!!!
     
  9. Nov 1, 2009 #8

    tiny-tim

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    I've told you … you have to use 9.8m.
     
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