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Rollercoaster& velocity problem Please Help

  1. Mar 19, 2006 #1
    Suppose the roller coaster in Fig. 6-41 (h1 = 38 m, h2 = 12 m, h3 = 20) passes point 1 with a speed of 1.40 m/s. If the average force of friction is equal to one third of its weight, with what speed will it reach point 2? The distance traveled is 35.0 m.
    m/s


    Please view atachment

    I dont even know where to start with this problem
     

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  2. jcsd
  3. Mar 19, 2006 #2

    Hootenanny

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    I'm afraid you problem isn't going to get answered ASAP, it takes a while for attachments to be approved. Perhaps if you described your problem?
     
  4. Mar 19, 2006 #3
    Or maybe you could host it in http://www.imageshack.us/ ...
     
  5. Mar 19, 2006 #4
    ok so there is a rollercoaster on the ramp and there are 3 diffeternt positions allocated (hi= 38m, H2=12m, H3=20m). Now according to the problem there is a friction force on the rollercoaster which amount to 1/3 of thr rollercoaster's weight. But they dont give the weight of the rollercoaster. Since there is a long distance between hi and h2 they want to know whats the velocity at point 2 ( which is btween H1 &2). My porblem is that i dont even know what equations i should be using because my prof always gave us similar problems in which mass was given. is so i would use :

    mgd=mgd2 + friction(distance)
     
  6. Mar 19, 2006 #5

    Hootenanny

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    That is the correct equation to use. Think about how else you could write the frictional force. What other equations do you know for friction?
     
  7. Mar 19, 2006 #6
    force of friction = coefficient of friction X normal force
    coefficient of friction mass(9.8)
    Force of Friction= coefficient of friction x mgcostheta
    with the other equation : friction =mgd/mgd2
    -----------
    distance
     
  8. Mar 19, 2006 #7

    Hootenanny

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    [tex]F_{friction} = \mu R = \mu mg[/tex]

    Substitute [itex]F_{friction} = \mu mg[/itex] into your equation and you will find that the masses cancel...
     
  9. Mar 19, 2006 #8
    umg=mgd/mgd2
    u=d/d2
     
  10. Mar 19, 2006 #9
    chazgurl4life,

    I think you did a mistake:

    umg=mgd/mgd2
    um = d/d2

    Only the masses in bold cancel and g.
     
  11. Mar 19, 2006 #10
    oh ok thanx
     
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