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Friction and velocity help

  1. Feb 9, 2007 #1
    1. The problem statement, all variables and given/known data
    An 85-N box of oranges is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward.

    2. Relevant equations
    Basically, f=ma


    3. The attempt at a solution
    Okay, this is what I attempted. I drew a FBD, so the eqn I came up with is
    fk - Fx = 0
    fk = Fx
    fn - w - Fy = 0
    fn = w + Fy
    And since fk = uk(fn)
    Then shouldn't uk = Fx/(w+Fy)?

    I know that's wrong because I didn't include the velocity. But I don't know how to integrate the velocity in there. If it's slowing down, that's only because of the friction force right? And at a constant rate just means the acceleration is zero....What other logic am I missing? Many thanks in advance.
     
  2. jcsd
  3. Feb 9, 2007 #2

    Doc Al

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    What you are given is not a velocity, but an acceleration. It's undergoing a constant acceleration, not constant velocity. The sum of forces in the horizontal (x) direction is not zero, since the box is accelerating.
     
  4. Feb 9, 2007 #3

    Hootenanny

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    Okay firstly, I need to correct your question. Notice the alteration is red
    If the box is slowing then it is undergoing an acceleration, which is the rate of change of velocity wrt time and has units of meters per second per second. I'm going to ignore your working, because I think you have a fundamental misunderstanding here.
    Correct
    This is simply wrong. If an object is slowing down how can the acceleration possibly be zero? Remember, as I said above that acceleration is the rate of change of velocity. It therefore follows that if the box is slowing down it has a non constant velocity (i.e. its velocity is changing) and hence the acceleration must be non-zero. Do you follow?

    EDIT: Dammit Doc :rolleyes:
     
  5. Feb 9, 2007 #4
    But the unit is m/s, isn't acceleration m/s^2?

    Edit: Okay nvm, I got it. Thanks for both the help!
     
  6. Feb 9, 2007 #5

    Hootenanny

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    You're correct, so I assume either you've incorrectly copied the problem or there has been a typo. However, you should have considered this yourself, the question as written is illogical; how can a box be slowing down and yet have a constant velocity?
     
    Last edited: Feb 9, 2007
  7. Feb 9, 2007 #6

    Doc Al

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    There's no typo and the units given in the problem statement are correct:
    Note that it says 0.90 m/s each second, which is equivalent to 0.90 m/s/s or m/s^2.

    (Of course you did leave out part of the problem statement: The question itself! Which I assume is to find the coefficient of friction.)
     
  8. Feb 9, 2007 #7

    Hootenanny

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    :redface:
    #Crawls and hides in shame#
    :blushing:
     
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