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Work and Force: Frictionless Rollercoaster

  1. Mar 19, 2006 #1
    Hi all, I'm having some problem in answering this question. If anyone could help me solve this problem, I will appreciate it very much...

    [​IMG]

    If a 12000 kg car starts at rest from Point A, calculate:
    a) the total energy of the system
    b) the speed of the car at point B
    c) the force that must be applied to bring it to a stop at point E
    d) the work done to bring it to a stop at point E

    I answered (a) by finding Et = (mv^2)/2 + mgy;

    And b and c by using the law of conservation of energy.

    But I don't know how to answer c. The work needed to stop the rollercoaster at y = 0 is basically equal to the kinetic energy. W=F*d*cos theta. The angle is zero. Therefore, F = W/d. However, the question does not provide the distance. If the distance is not given, how could we find the force required to stop the roller coaster? Or is this question just simply wrong? Please do leave your thought even if you can't answer this question since it wouldn't suprise me if this question is wrong. Speaking from my experience in taking this distance-ed course, some other questions are not completely stated...


    Thanks...

    James
     
    Last edited: Mar 19, 2006
  2. jcsd
  3. Mar 19, 2006 #2
    The information provided is definitely insufficient.

    There is a double headed arrow indicating distance to point E from the point at which y=0 initially. So it is probably a misprint in the diagram.
    Also they ought to say in which direction the force is being applied.
     
  4. Mar 19, 2006 #3
    Thanks arunbg, that's what I thought... I'm glad I'm not the only one that thinks the information is insufficient.

    Just to confirm this, does anyone else have any thought on this?
     
  5. Mar 19, 2006 #4

    Hootenanny

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    Yes, the information here is incomplete. Also, I think it would make more sense to ask question (d) before question (c) becuase you need work done to calculate F, so you work out (d) while answering (c) anyway.
     
  6. Mar 19, 2006 #5
    Hootenanny: Yes, I think I'd just mention that it's impossible to do (c) because the distance is not given. Thank you for your thought. :=)
     
  7. Mar 20, 2006 #6
    Hi all,

    I just thought of something. Please let me know if it makes sense.. Would the force have the same magnitude of Fg since the track is frictionless and no energy is loss?

    James
     
  8. Mar 20, 2006 #7
    The path is an ellipse. Do you think about something that happens when the path is a curve?
     
  9. Mar 20, 2006 #8
    Well, I lurve energy so much LOL.

    My way would be if I looked at this:

    a) I would just find the GPE at Point A since the car is not moving initially to get the total energy of the system
    b) Just find your GPE at that point, subtract from total energy to get the KE at that point and yada yada
    c) Now, you said this track was frictionless so it has an acceleration and velocity at Point E. I would say that Point E is 0 GPE so find it's KE at that point. When WE apply a force, it is non-conservational. So obviously, you CANNOT SOLVE c), without solving for d) first...
    d) The Work(other) is work done by non-conservative forces, this includes friction and outside forces (such as an applied force).
    [tex]W_{other} - \triangle U = \triangle K[/tex]
    or
    [tex]W_{other} = \triangle E[/tex]
    Now that we have this, you can.. 'kinda' solve for c) although I ain't sure how the heck you are gonna get distance there.. It maybe that there's a different way that you don't need distance at all :-| I'll think about this one..
     
  10. Mar 20, 2006 #9
    Hmm, not especially.. Friction is based on it's force normal... So not really, might be lucky but NAH o_O

    Mass influences it as F=ma so maybe rewrite as F=mv/t and get rid of somestuff o_O
     
  11. Mar 20, 2006 #10
    Yeah, I know the path is an ellipse.. This is just a hopeful thinking, since I cannot find any other way to solve it. And I know that most likely, my idea is wrong.

    But I was just thinking since at Point A, *maybe* the only force that is causing the rollercoaster movement is the force of gravity, then *maybe* the amount of force required to stop the coaster is the same as that..

    But then again, there is a normal force involved acting on an angle when it's sliding down.. So perhaps my idea is just wrong..

    What do you think?
     
  12. Mar 20, 2006 #11
    Exactly, F = m (vf-vi)/t but there's no t is mentioned :-(
     
  13. Mar 20, 2006 #12
    I just thought of like impulse and momentum but that would be taking it too far as it should be just.. energy o_O
     
  14. Mar 20, 2006 #13
    Exactly. If noone thinks my idea makes sense, I think I'll just give up. Lol. I'll write some comments mentioning this question is unworkable.
     
  15. Mar 21, 2006 #14

    Hootenanny

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    There isn't enough information here to solve the question. You are quite right, you could use the impulse to find the force and then use kinematic equations to find the distance. However, you are not given sufficent data to make this possible.

    As arunbg said I think the distance E is missing from the diagram.

    - Hoot :smile:
     
  16. Feb 10, 2009 #15
    I am sorry, but to all you physics mentors, ILC I am sure would know all about their questions. This one has enough information. In fact, questions (c) and (d) are the easiest. (D) is the easiest question ever! I think what you might be confused about is that the diagram, after the last hump, has a horizontal space of 7 meters. It is over this space that the roller coaster must be stopped. I am surprised none of you could figure this out. I did, very simply and easily.
     
  17. Mar 2, 2009 #16
    could you explain how you would go about doing c and d?

    I understand that the change in distance will be the last 7 meters. so obviously we will deduce a value for "work" to answer this question. I'm confused on how to consider the forces though. I know there is no friction...what about an applied force? In addition, I'm assuming theta = 0 degrees. In this case is the force simply fg? therefore work = 12000x9.8x7

    thanks
     
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