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Dynamics-Coefficients of Kinetic friction

  1. Jul 8, 2008 #1
    1. The problem statement, all variables and given/known data

    A student is pushing horizontally on a table (m=16 kg) to move it across a horizontal floor. The coefficient of kinetic friction between the table and the floor is 0.61.
    a) Determine the magnitude of the applied force needed to keep the table moving at constant velocity.
    b)If the applied force were 109N and the table were to start from rest, how long would the table take to travel 75 cm?

    2. Relevant equations

    Fk =Uk |Fn|

    3. The attempt at a solution
    I figured out part a) by substituting the mg for the Fn of the second equation. This helped me get the equation Fapp = Uk mg, to which I inputed all the correct values. I obtained an answer of 96 N. However, I do not know what equation to use for part b) or how to solve it. Any suggestions would greatly be appreciated. :)
  2. jcsd
  3. Jul 8, 2008 #2


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    Hi shakabra,

    To figure out part b, think about what you did for part a. There you set the forces equal to each other to solve for F. Why specifically did you do that?

    For part b, the forces won't be balanced. So what will be different about the motion?
  4. Jul 8, 2008 #3
    Well for part a) I set the forces equal to each other only because I was using the equation Fk=Uk |Fn| as base to which I substituted values until I received a suitable equation.
    For part b), I'm not too sure what would be different about the forces. The question says that the student is pushing horizontally on a horizontal floor. Is the applied force in part b) perpendicular to the surface?
  5. Jul 8, 2008 #4


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    This equation gave you the frictional force, and the problem asked for the applied force. In this case, you correctly said that the frictional force equalled the applied force.

    What in the problem indicates that these two forces should be equal?

    In part b, these two forces are not equal; what happens if unequal horizontal forces act on the object?
  6. Jul 8, 2008 #5
    If the two forces are not equal then that means that the object moves.
    But knowing that, I still don't know what I'm suppposed to do ...
  7. Jul 8, 2008 #6


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    Well, in part a, when the forces are equal, the object is also moving. However, it is moving at constant velocity. When the forces are unequal, there is an acceleration.

    In your relevant equations, you have [itex]F=ma[/itex], but to be more precise we should say either

    F_{\rm net} = m a \mbox{ or } \sum F= m a

    (for 1D motion). In part a, you know a=0, so the sum of the forces equal zeros; that's why their magnitudes must be equal.

    In part b, you know the individual forces, so you can find the acceleration using [itex]\sum F = ma[/itex]. Once you have the acceleration, you can use kinematics to find time of travel.
  8. Jul 8, 2008 #7
    I got the answer!! The time of travel is 1.3s.
    Thank you so much for your help alphysicists! :)
  9. Jul 8, 2008 #8


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    Great! and I'm glad to help.
  10. Aug 3, 2008 #9
    i dont understand the concept could u plz help me out by answering the question because i have been trying this question for quite some time now.
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