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Conservation of Energy for a roller coaster

  1. May 5, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-5-5_18-19-7.png


    If it can be assumed that the car has the same speed at points A and E, which of the following statements is true?
    a. The net work done in this system is 0
    b. The net work done in this system is positive
    c. The net work done in this system is negative
    d. The net work done in this system is positive for the roller coaster but 0 for the slope
    e. The net work done in this system is 0 for the roller coaster but positive for the slope

    2. Relevant equations
    We know that in this scenario, energy is completely cosnerved. Therefore, the total energy at A equals the total energy at E.

    3. The attempt at a solution

    It can't be (d.) or (e.) because the work is the same for the roller coaster and slope due to Newton's Third Law.

    I think it's (a.) since the net change in energy is 0, since the cart returns to position E with the same potential energy.

    Correct answer is (b.), I don't see why.
     
  2. jcsd
  3. May 5, 2017 #2

    kuruman

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    What does the work-energy theorem have to say about all this?
     
  4. May 5, 2017 #3
    The energy in A is different to the energy on E since E is at a lower height. It should mean that the car gained some kinetic energy at E. I think the statement is wrong because if the car has the same speed on both points, it should have same kinetic energy that also means same height in a conservative system of forces. On the other hand remember that even with energy conservation the changes in energy are due to work.

    EDIT:
    I think that as the diference in height is small. the asumption is to consider only the change in potential energy to compute the work done.
     
  5. May 5, 2017 #4

    kuruman

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    Which energy are you talking about? Kinetic, potential or mechanical?
     
  6. May 5, 2017 #5
    Potential sorry. And kinetic should be diferent also so mechanical energy is the same.
     
  7. May 5, 2017 #6

    kuruman

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    But the assumption is that the speed is the same at points A and E ...
     
  8. May 5, 2017 #7
    Yes you are correct, that is the assumption!
     
  9. May 5, 2017 #8

    haruspex

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    It is a standard rule in logic that if you start with a falsehood you can prove anything.
    How can it be frictionless, no change in KE of the cart, loss of PE of the cart? Where has the energy gone?
    Air drag, or is "frictionless" supposed to rule that out too? The slope itself is not fixed?

    Bizarre question.
     
  10. May 5, 2017 #9

    kuruman

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    Probably air drag or some such dissipative force; I cannot think of another mechanism that would convert PE to something other than KE.

    It is not a well-constructed question for testing the understanding of the work-energy theorem. Choices d and e are confusing. I think the words "in the system" must be removed from them if the choices are to make sense. I assume that by "system" is meant "roller coaster + track".
     
  11. May 6, 2017 #10
    It is a weird question. But I guess the central point is that if we were to ignore the friction-less part, then the change in kinetic energy is negative, since the car would've had greater speed since it is at a lower height. So some kinetic energy was lost.
     
  12. May 6, 2017 #11
    Thanks everyone for their input on this question!
     
  13. May 6, 2017 #12

    haruspex

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    No, the cart is at the same speed, so no (net) KE was lost. (PE was lost.) What does that tell you about the work done on the cart?
     
  14. May 6, 2017 #13

    kuruman

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    The choices are contingent upon the statement
    You cannot assert that the change in kinetic energy is actually negative because that changes the question very radically.
     
  15. May 6, 2017 #14
    Wait but this is how I understood it:

    Point A had potential energy, and [K] kinetic energy
    Point E has [U2] potential energy, which is less than .
    If the system (cart and track) is energy conserving, then the decrease in must be transferred to an increase in [K]
    Therefore, the kinetic energy at E should be [K2], which is more than [K]

    But the kinetic energy stayed the same. So a dissipating force, (probably air drag) is where the extra kinetic energy ([K2] - [K]) went.
    Wouldn't this be a loss of kinetic energy?
     
  16. May 6, 2017 #15
    Ohhh are you guys saying:

    The potential energy at E is less than the potential energy at A?
    (Without reference to the kinetic energy)

    So thus the change in potential is negative, therefore the work done is positive?
     
  17. May 6, 2017 #16

    haruspex

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    It s certainly zero for the cart, but what it is for the system depends on how the system is defined. The question does not specify. Is it cart plus the whole of the Earth, including the track, or is it cart plus track, separated at some boundary from the rest of the Earth?
    When the cart loses PE, what is doing work on what?
     
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