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Kinetic Friction and box

  1. Oct 7, 2008 #1
    1. The problem statement, all variables and given/known data
    A 40.0kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130 N. The coefficient of friction between box and floor is .300.
    Find the change in kinetic energy of the box.


    2. Relevant equations
    (delta)K=K(final)-K(initial)
    K(final)=1/2mv^2


    3. The attempt at a solution
    Work=650. J
    (change)Internal Energy=588 J
    Fk=118

    How do you find the change in Kinetic energy without the velocity?
     
  2. jcsd
  3. Oct 7, 2008 #2
    What does the work-energy theorem state?
     
  4. Oct 7, 2008 #3
    When work is done on a system and the only change in the system is in its speed,
    the work done by the net force equals the change in kinetic energy of the system.

    Wnet = Kf - Ki = deltaK

    In this situation Ki = 0, Kf = 1/2mv^2

    Right?
    This is a classmate of mine and we are working on this together.
    Kevin
     
  5. Oct 7, 2008 #4
    That's correct. So you have the distance over which the force acts. What is the net force in this situation?
     
  6. Oct 7, 2008 #5
    It is the Work force of 650J - the frictional force of 588J to get 62J?
    Kevin
     
  7. Oct 7, 2008 #6
    That looks right to me.
     
  8. Oct 7, 2008 #7
    Now we will try and work on her other one with the skier that she posted and we
    should be done.
    Thanks for the help,
    Kevin
     
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