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Homework Help: Work and energy roller coaster question

  1. Oct 29, 2008 #1
    1. The problem statement, all variables and given/known data
    A 250kg roller coaster passes a point A(located at the top of a hill of height 18.0m) at 12.0 m/s.

    What is the speed of the roller coaster at point B located at the bottom of the hill if 8500J of energy is transformed to heat during the journey?

    2. Relevant equations
    ΔE = heat

    3. The attempt at a solution
    First I wrote down what I knew from reading the question:
    m = 250kg
    g = 9.0 m/s (squared)
    vi = 12.0 m/s
    hi = 18.0m
    hf = 0m
    heat energy = 8000J

    Someone showed me how to do this question and they said that since the question only tells you what the mass, initial velocity, initial height, final height, and heat energy are, that you can just use the equation for heat (ΔE=heat).

    ΔE = heat
    (Epf-Epi) + (Ekf-Eki)=heat
    Epf + Ekf = Epi + Eki + heat
    mghf + (1/2)mvf(squared) = mghi + (1/2)mvi(squared) + heat

    After putting in the numbers and isolating vf, I got a final velocity of about 23.7 m/s. But I was just wondering is this correct, or do I need to find the work and use the equation "Wnc=ΔE" and then use that and my other findings to isolate and find vf? Because this question is out of 10 marks so I was just wondering whether there was more to this question (in the attempt to find vf) calculation wise or was what I did above enough or correct to find the final velocity?
  2. jcsd
  3. Oct 29, 2008 #2
    In this problem mechanical energy is not conserved since it is lost to heat... so you need to find the remaining enegy left to give the roller coaster a velocity.
    Find the potential energy stored in the roller coaster at its initial point:
    it will be easiest if you set the lowest point (hf) to be zero.
    now if there were no friction, air drag etc all the mechanical energy would be conserved and its Ei would be the same as its Ef...
    But it isnt!! so all you have to do is account for the energy lost as heat...

    Ty rake-mc for correction
    Last edited: Oct 30, 2008
  4. Oct 30, 2008 #3
    You should probably re-phrase that such that you don't confuse him. Energy is always conserved. In this case, mechanical energy is not conserved, but the total energy in the system remains constant. If it didn't you wouldn't be able to solve this question with only this information given.
  5. Oct 30, 2008 #4
    so would it be:
    ΔE = -heat
    (Epf - Epi) + (Ekf - Eki) = -8000J
    (mghf-mghi) + ((1/2)mvf(squared) - (1/2)mvi(squared)) = -8000
    (mg(0) - mghi) + ((1/2)mvf(squared) - (1/2)mvi(squared)) = -8000
    -mghi + ((1/2)mvf(squared) - (1/2)mvi(squared)) = -8000
    -(250)(9.80)(18.0) + (1/2)(250)(12.0)(squared) - (1/2)(250)(vf)(squared) = -8000
    (-44100) + 18000 = -8000 + (1/2)mvf(squared)
    (-26100) + 8000 = (1/2)(250)(vf(squared))
    -36200/250 = vf(squared)
    -144.8 = vf(squared)

    but then how would you get rid of the negative so that you could square root the vf to get the final velocity?
  6. Oct 30, 2008 #5
    Sorry, that's really hard to read haha,

    This is how I'd go about it:

    First I write out all the types of energy that we have (omitting heat initially) on both sides like this:

    [tex] \frac{1}{2}mv_1^2 + mg\Delta h = \frac{1}{2}mv_2^2 + mg\Delta h [/tex]

    This is the conservation of mechanical energy for your system. However some of the mechanical energy is transformed into heat energy, therefore we can re-write it like this:

    [tex] \frac{1}{2}mv_1^2 + mg\Delta h = \frac{1}{2}mv_2^2 + mg\Delta h +8500J [/tex]

    Now, we know that the final height is zero (from the co-ordinate system we define). Therefore we can cancel [tex] mg\Delta h [/tex] from the right hand side so we have:

    [tex] \frac{1}{2}mv_1^2 + mg\Delta h = \frac{1}{2}mv_2^2 +8500J [/tex]

    Re-arrange for V2 and solve.

    EDIT: Also, in your class, have you been taught to take acceleration due to gravity as 9.0ms-2? Most physics classes teach it as 9.8 or 9.81ms-2 in some cases it's even 10.0ms-2 but I've never heard of it as 9.0ms-2.
  7. Oct 30, 2008 #6
    Thank you for all your continued help and support for this question. After looking at your example I think I now have a deeper understanding as to what the question is asking and now know what to do for this question.

    As for what our class uses for the acceleration of gravity, we use 9.80 m/s(squared).
  8. Oct 30, 2008 #7
    Okay good, no worries. The most important thing is that you're not just getting the answers, but you're learning how and why it all works.
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