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Finding Total Energy of a system

  1. Jun 18, 2009 #1
    Alright, the problem is fairly simple actually; however, I've seen a few methods to solve this that have kind of confused me.

    The questions is:

    Consider a frictionless roller coaster with a mass of 12,000kg. If the coaster starts at rest at point A, which is 95m above the ground, calculate the total energy of the system.

    My attempt at a solution:

    Since the coaster is at rest at point A, the total energy will simply be equal to the potential energy of the system;

    ET = EK + EP

    ET = [tex]\frac{mv^{2}}{2}[/tex] + mgh

    ET = [tex]\frac{12000kg * 0m/s^{2}}{2}[/tex] + (12000kg × 9.8N/kg × 95m)

    ET = EP

    ET = 1.1×107 Joules

    Therefore, the total energy of the system is 1.1×107 Joules.

    Now, the question asks:

    Calculate the total speed of the coaster at point B which is 65m above the ground (30m below point A).

    This is where I got confused. I was told that I could just find the potential energy of the system at this point and subtract it from the total energy found at Point A and use the kinetic energy formula to solve for speed. But I don't understand why I would subtract the energy at point A from the energy at point B. I figured I could just use the total energy formula and substitute in the new height (65m) and isolate the speed variable and solve:

    ET = [tex]\frac{mv^{2}}{2}[/tex] + mgh

    v = [tex]\sqrt{\frac{E_{T} * 2 - 2gh}{m}}[/tex]

    From this equation, I'm getting approximately 1.3m/s. Now that seems a bit low to me. If someone could clarify, or at least let me know if I'm on the right track that would be great. Thanks.
     
  2. jcsd
  3. Jun 18, 2009 #2
    You're missing an m.

    v=√((ET - mgh)*(2/m))

    A simpler approach to follow would be to just move your plane of reference, to avoid getting mixed up in the math.
    If you define U=0 for h=65m, you have all the potential energy (Just 30 meters' potential, though) converted into kinetic energy, which simplifies the math considerably.

    U = mgh, with h being measured relative to the new plane of reference, you only get a factor of 30 (m).

    mgh = ½mv²
    -->
    v² = 2gh (This is a result you should remember, it'll come up a lot)
     
    Last edited: Jun 18, 2009
  4. Jun 18, 2009 #3
    Ooooh! Right. It's the same as using this formula:

    vf2 = vi2 + 2a[tex]\Delta[/tex]d

    vf2 = (0 m/s)2 + 2(9.8 m/s2)(30m)

    vf = [tex]\sqrt{588 m^{2}/s^{2}}[/tex]

    vf = 24.3 m/s

    Ok, so as the coaster decends to Earth's surface, it is losing potential energy, which is being transformed into kinetic energy.

    But one more thing I don't understand is that if I were to plug this value for speed into the kinetic energy formula, I get a pretty large number, larger than the potential energy of the system. This is not what I expected because the coaster has only dropped 30m, which is not even half-way. At half-way (47.5m), shouldn't it be expected that the potential and kinetic energies are equal? (First law of thermodynamics -- Total energy is constant in a system) Therefore,

    ET = EK + EP

    And I've already calculated the total energy of the system to be equal to the coaster's potential energy at point A; 1.1 * 107 Joules

    Substituting the value for speed found in the preceeding formula into the kinetic energy formula I get:

    EK = [tex]\frac{1}{2}[/tex]mv2

    EK = [tex]\frac{1}{2}[/tex](12000 kg)(24.3 m/s)2

    EK = 3.5 * 109 Joules

    Clearly this is much larger than the total energy of the system.
     
  5. Jun 18, 2009 #4
    Be careful of that line of reasoning. The 'total energy' you started with is COMPLETELY arbitrary.

    But with 'height' energy, then there's a linear relationship between the height descended and the kinetic energy gained.

    At half-way, yes, you'd have half of the 'height' energy converted to kinetic energy, and you'd have half of the 'height' energy left, making U=Ek, because you've started with Ek=0

    You made a calculation error.

    Ek = ½mv² = ½*(12000 kg)*(24.3 m/s)² = 3542940 J = 30/95 * Ug i
     
  6. Jun 19, 2009 #5
    Ok I've made the correction; however, I'm still a bit confused with regards to saying that the total energy of the system is completely arbitrary.

    In what sense is it arbitrary? I was under the impression that that value was pretty significant.
     
  7. Jun 19, 2009 #6
    Only the difference between the initial and final state is significant. In choosing the plane of reference, you make an arbitrary choice. Sure, some are more convenient than others, but none are wrong.

    You could have set your plane of reference at the top just as well, the only things that matter are the invariant sizes, the differences in potential energy.
     
  8. Jun 19, 2009 #7
    Ooooh I see. That makes a lot of sense.

    Then does that mean that the total energy that I found isn't in fact the total energy?

    Would the total energy then be the potential energy I found at point A + the kinetic energy I calculated at point B?
     
  9. Jun 19, 2009 #8
    No. You define the total energy in reference to a certain plane of reference if you're using potential energy.
    'Total energy' is conserved, in relation to that specific plane of reference.
    Take for example the following case:
    First, we set our plane of reference at the top of the roller-coaster (95m above ground-level).
    Etot i = Ug + Ek
    Etot i = 0

    If we let it slide all the way down, we'll have kinetic energy, and negative 'height' energy:
    Etot f = -mg*h + ½mv²
    Since this has to equal 0, you can quickly see the following holds true: v² = 2gh
    You would get the exact same result regardless of your plane of reference since h is the height descended, and is invariant regardless of how you choose to look at things.

    Potential energy is a useful construct, but it is not necessary, it just makes things A LOT more simple.
    I think of it as the work a conservative force would perform on the object, translating it from its initial position to the plane of reference at an infinitesimal constant speed.
     
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