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Kinetic and Gravitational Energy Question

  1. Jul 15, 2011 #1
    A dim but resourceful student has a summer job in a factory moving boxes. The 50 kg student finds that if he runs at a speed of 5.8 m/s and then dives head first into the boxes, he can move them across the floor. If the boxes have a mass of 65 kg and the force of friction between the boxes and the floor is 382 N, how far will he move the boxes after hitting it six times?

    How am I suppose to solve this question?

    m of student = 50 kg
    m of boxes = 65 kg
    v = 5.8 m/s
    force of friction = 382 N
     
    Last edited: Jul 15, 2011
  2. jcsd
  3. Jul 15, 2011 #2
    How do you relate force with work? Can you do this to find the work done by the frictional force?
     
  4. Jul 15, 2011 #3
    Work is done whenever a force causes an object to move...but I don't understand how the force of friction is involved in the process of finding the distance for this question (what formulas are used?).
     
  5. Jul 15, 2011 #4
    What's the formula for work?
     
  6. Jul 15, 2011 #5
    w = F x d
     
  7. Jul 15, 2011 #6
    You have the force (F), and you need to find the distance (d). Do you know where w could come from?
     
  8. Jul 15, 2011 #7
    Gravitational potential energy?
     
  9. Jul 15, 2011 #8
    But you don't have any information on height - you have information on the student's velocity. This should tell you the type of energy you need to use for w.
     
  10. Jul 15, 2011 #9

    SteamKing

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    I think the floor might be level, so the gravitational potential energy is probably not changed when the boxes are moved.

    Say, what about that fellow who runs and dives head first into the boxes? Do you think there might be something wrong with him?
     
  11. Jul 15, 2011 #10
    Kinetic energy? So you have to find the Kinetic energy of the student? If so then the Kinetic energy would be 145 J.
     
  12. Jul 15, 2011 #11
    Yes, you can equate it with the work done by the frictional force to find the distance (d). This works because the kinetic energy is dissipated by the frictional force.
     
  13. Jul 15, 2011 #12
    So I should now use the formula d = W/F, where F is the force of friction (382 N)?
     
  14. Jul 15, 2011 #13

    gneill

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    Staff: Mentor

    A point for the helpers in this thread to consider: When the student dives headlong into the boxes, will this be considered an elastic or an inelastic collision? Will it make a difference to the outcome? If so, how? :smile:
     
  15. Jul 15, 2011 #14
    I assume an elastic collision so the kinetic energy is constant in the collision. Of course, this won't be possible in real life, but the question doesn't mention anything either!

    @NeomiXD - Yes the d you get should be for one dive. You can multiply your answer by 6 to get the total distance.
     
  16. Jul 15, 2011 #15

    gneill

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    In an elastic collision the KE may be conserved but it will be split between the box and the student...
     
  17. Jul 15, 2011 #16
    I was just wondering about the mass of the boxes (65kg). Doesn't that affect the distance too?
     
  18. Jul 15, 2011 #17
    Yes, but then again the question doesn't say anything about the velocity of the student afterwards. I'd assume the student comes to rest so that all kinetic energy is with the boxes (hence covering maximum distance). I approach these questions with such "ideal circumstances" in mind as discussed, because most of the time that is what is intended.
     
  19. Jul 15, 2011 #18
    The kinetic energy of the student is completely transferred to the boxes, so they have energy equal to [itex]\frac{1}{2} m v^{2}[/itex], where m is the mass of the student, and v is the velocity.
     
  20. Jul 15, 2011 #19

    gneill

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    The student would only come to rest if his mass were equal to that of the box.

    Simplifying assumptions are only useful if they they don't break physical laws :smile:. To quote Albert E., "Everything should be made as simple as possible, but no simpler".
     
  21. Jul 15, 2011 #20
    True - I didn't consider this. So the correct way to proceed would be to first find the new velocity of the box from the collision, right?
     
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