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Acceleration (newton's law)

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

    A bureau rests on a rough horizontal surface ( mk=0.40, ms=0.50). A constant horizontal force, just sufficient to start the bureau in motion, is then applied.The acceleration of the bureau is:

    2. Relevant equations

    F= ma

    3. The attempt at a solution

    The answer is 0.98 m/s^2 but I'm not sure why. I thought you would only use the kenetic friction since it just starts to move.
     
  2. jcsd
  3. Nov 1, 2007 #2

    Dick

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    You use the static friction to get the applied force. Once it's moving the applied force doesn't change, but now use kinetic friction in your force balance.
     
  4. Nov 1, 2007 #3
    how can you get the applied force using the coefficient of static friction?..Coefficient of static friction= Fs/Fn
     
  5. Nov 1, 2007 #4

    Dick

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    Fn=mg. Fs=0.5*mg. To start it moving the applied force must be equal to Fs.
     
  6. Nov 1, 2007 #5
    ohk..I see..so grouchy's answer is right, although the method is not
     
  7. Nov 1, 2007 #6

    Dick

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    He didn't have a method. Hopefully now he does.
     
  8. Nov 1, 2007 #7
    ok, so you take the fs and subtract the fk? right?
     
  9. Nov 1, 2007 #8

    Dick

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    Why the question marks? fs is the force pushing the cabinet - fk is the force acting against the push. Use the total force on the cabinet to compute it's acceleration.
     
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