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Monkey and sled - conservation of energy problem

  1. Jun 20, 2006 #1
    I'm working on the following problem:

    a monkey is strapped to a sled and both are given an initial speed of 4.0 m/s upa 20 degrees inclined track. The combined mass of monkey and sled is 20 kg, and the coefficient of kinetic friction between the sled and the incline is 0.20. How far up the incline do the monkey and sled move?

    Using the conservation of energy equation, I knowthat
    KEi + PEi + Wnc = KEf +PEf

    Since both PEi and KEf are 0, the equation is as follows:

    KEi+Wnc=PEf

    how do i find Wnc?
     
  2. jcsd
  3. Jun 20, 2006 #2

    Hootenanny

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    First just a slight correction;
    This should read; KEi + PEi = KEf + PEf + Wnc. The intial kinetic energy will be converted into potential energy and work, thus work and final potential energy should be on the same side of the equality. As you correctly say, the inital potential energy and final kinetic energy drops out leaving you with; Wnc = KEi - PEf. Now, you know the intial kinetic energy, you must next calculate the change in potential energy. For this you need to calculate the height. HINT: Trigonomentry. Can you go from here?
     
  4. Jun 20, 2006 #3
    i know that KEi is 1/2 x 20 kg x 4.0 m/s^2, and PEf is 20kg x 9.8 m/s x sin(20) x d

    right?
     
  5. Jun 20, 2006 #4

    Doc Al

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    Start by finding the friction force acting on the sled. (Identify all the forces acting on the "monkey/sled" system.)
     
  6. Jun 20, 2006 #5
    but d is still unknown
     
  7. Jun 20, 2006 #6
    F=u Fn (u being the coefficient of friction), F being the normal force
     
  8. Jun 20, 2006 #7

    Doc Al

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    If you set up your energy equation properly, "d" will be the only unknown. To find d, solve that equation.
     
  9. Jun 20, 2006 #8

    Hootenanny

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    That is correct, so now you have;

    [tex]dR\mu = 160 - 196d\sin(20) \Leftrightarrow d(R\mu + 196\sin(20)) = 160[/tex]

    Can you go from here?
     
  10. Jun 20, 2006 #9
    normal force = mass x acceleration due to gravity?
     
  11. Jun 20, 2006 #10

    Hootenanny

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    Careful, the normal force always act perpendicular to the surface. Gravity in this case is not acting perpendicular to the surface.
     
  12. Jun 20, 2006 #11
  13. Jun 20, 2006 #12

    Doc Al

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    Right. So what's the normal force (Fn)?

    Once you've got the the friction force, use it to express the "work" done against friction.
     
  14. Jun 20, 2006 #13
    i don't know how to find the friction force. force = ma, but how can i find the acceleration?
     
  15. Jun 20, 2006 #14

    Hootenanny

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    For kinetic friction, the frictional force is simply given by;

    [tex]F = \mu R[/tex]

    Where R is the normal reaction force
     
  16. Jun 20, 2006 #15
    how do you find the normal reaction force?
     
  17. Jun 20, 2006 #16

    Hootenanny

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    As Doc Al suggested draw a free body diagram of all the forces acting.
     
  18. Jun 20, 2006 #17
    anyone there?
     
  19. Jun 20, 2006 #18
    oh sorry, comp glitch
     
  20. Jun 20, 2006 #19

    Hootenanny

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    Draw a free body diagram. HINT: The sled is only moving parallel to the incline. Resolve all forces so that they are either parallel or perpendicular to the inclined plane.
     
  21. Jun 20, 2006 #20
    i'm not sure what other forces are acting. force of gravity, you said no.

    force of sled monkey?
     
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