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Conservation of Energy problem

  1. Jun 2, 2007 #1
    1. The problem statement, all variables and given/known data

    A 4.0X10 to the power of 4 kg roller coaster starts from rest at point A. Neglecting friction, calculate the potential energy relative to the ground, its kinetic energy, and its speed at points B, C and D.

    Height A=54m
    Height B=15m
    Height C=47m
    Height D=35m


    2. Relevant equations

    Eg=mg triangle (height)- Triangle=multiply height
    Ek=1/2mv square

    3. The attempt at a solution

    I found the answer for the total potential energy from A to D which is 5.9X10 to the power of 7 J. I got height, mass, and total potential energy.
    I need help finding the kinetic energy and the speed for points B, C, and D with a step by step solution.
     
  2. jcsd
  3. Jun 2, 2007 #2
    Knowing that gravitational field is conservative, the potential energy is:

    Eg=mgh.

    So if the heights of the points are given relative to the gorund, then the potential energy in every point is the wrote above.

    Knowing that the total energy is:

    Et=Ek+Eg

    In the other side, I have calculated the potential Energy in A and D relative to ground, and in A relative to D (knowing that the gravitational field is conservative, only the initial and final point matters, so the h used in that case is h=hA-hD). And in none of the results I have reproduced your "5.9X10 to the power of 7 J".

    Maybe I didn't understood what did you meant with the "total potential energy from A to D" due the english is not my mother languaje, but I thought it was the potential energy of A relative to D.
     
    Last edited: Jun 2, 2007
  4. Jun 2, 2007 #3

    danago

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    Gold Member

    The law of conservation of energy states that energy is never lost or gained, but can be transformed into another form of energy. Since we are neglecting friction and other losses of energy, we can assume that all energy remains in the system as kinetic and potential energy.

    From the law of conservation of energy, you know that at any point (A, B, C, D), the total amount of energy in the system will be the same, but with different amounts of kinetic energy (Ek) and potential energy (Ep).

    The first step is to calculate how much energy (Ek+Ep) is actually in the system. You can consider any of the 4 points to calculate this. However, at only one of these points do you have sufficient information to do so, since you need to know the height and velocity.

    Think about which point you have enough information to calculate the total energy in the system, and then see if you can use that amount to find the velocity at the other points.

    EDIT: ^^ ahh beat me to it :p
     
  5. Jun 2, 2007 #4
    I need help solving its kinetic energy, with a step by step solution and its speed at points B, C and D. I knw Ek=1/2m square but iunno how to calculate that part even with mass, height for each point, and gravational potential energy. I also want to knw the speed for points B-D.
     
  6. Jun 2, 2007 #5
    Think what happens if the kinetic energy is zero in one point, use the law of conservation of energy.
     
  7. Jun 2, 2007 #6
    I dun tink i got the correct answer. Ek=1/2mv square
    5.9X10 to the power of 7=1/2(68kg)v square.
    I need a step by step solution for kinetic energy and speed from point B to D.
     
  8. Jun 2, 2007 #7

    danago

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    Gold Member

    Well at any point, the total energy in the system is given by:

    [tex]E_t=E_k+E_p=\frac{mv^2}{2}+mgh[/tex]

    At point A, you know v=0, h=54 and m=4x104, so you can calculate the total energy. Do that first.

    This total energy is the same for any point on the track. So lets look at point B now. The total energy, again, is given by:

    [tex]E_t=E_k+E_p=\frac{mv^2}{2}+mgh[/tex]

    Now this time, we know h=54 and m=4x10, but we dont know v. However, because of conservation of energy, Et will be the same as at A, so substitute these values in, and solve for the velocity at point B.
     
    Last edited: Jun 2, 2007
  9. Jun 2, 2007 #8
    Well i never used that equation but i give it a try.

    mv square/2 +mgh
    40000(0) square/2=0
    40000(9.81)(54m)=2.1 X 10 to the power of 7.
    Now wat should i do?
     
  10. Jun 2, 2007 #9
    Now that you know the total energy, you can apply that to other point (for example, B).

    Knowing that Ep=mgh, use the height of B to calculate Ep, then you have Et (done by yourself), soy you can have v with some algebra.
     
  11. Jun 2, 2007 #10
    Oh alrite i found the answer of 1.5 X 10 to the power of 7. From A subtract B using the Ek=Eg formula.
    Now i need to find the speed for points B, C, and D.
     
  12. Jun 2, 2007 #11
    Well, you have there. If the EpA-EpB=1'5·10^7 J, then that is the potential Energy that became kinetic, so you have that...

    EpA-EpB=EkB=m(vB)^2/2

    For a point x:

    EpA-Epx=Ekx=m(vx)^2/2
     
  13. Jun 2, 2007 #12
    Now i have potential and kinetic energy. I need to find the speed of point B to D. I need a step by step solution with the formula.
     
  14. Jun 2, 2007 #13
    See that in the Ek formula (Ek=mv^2/2) you have the speed for each Ek (and you have that m=4·10^4).
     
  15. Jun 2, 2007 #14
    So your saying i have that m=4.0X10^4. I got the potential energy for A which is 2.1X10^7, B which is 5.9X10^6, C which is 1.8X10^7, and D which is 1.4X10^7 and B's kinetic energy is 2.1X10^7-5.9X10^6=1.5X10^7. So how do i find the kinetic energy for C, D, and E. I also need to find the speed for point B, C, and D. I want to see a step by step solution.
     
  16. Jun 2, 2007 #15
    Exactly, now that you know that EkB=1'5·10^7 an that Ek=m·v^2/2, you can write:

    1'5·10^7=m·v^2/2. Is an equation where you know m (4·10^4), so simply solve v in that equation.

    For the other points, you have that the Ek is the difference between Et=EpA and the Ep in that point (same that you did in B). This is because the equation that danago worte.

    With every Ek you can solve the speed like you can do in B.
     
  17. Jun 2, 2007 #16
    I found v=27 for B but i cant find the kinetic energy for C and D.
    As u told me i need EGa-EGb=EKb but i dun seem to have the correct answer for C and D.
     
  18. Jun 2, 2007 #17
    Like I said before, the gravitational field is conservative. That means that no matter the path, only the initial and final point for the Ep. You have the Ep of every point. Now, the same thing that you made for B you can do for other points. EkC=EpA-EpC, and so on. Think that EpA is the Et, so you can calculate every point like you make in B, for example D, no matter if it went through B or C before.

    Having the Ek in C, you can solve the v in C, like you did in B (its v=28 (remember to round well, 27'6 is rounded to 28), like you have said).
     
  19. Jun 2, 2007 #18
    well i did 1.5X10^7 J=1/2mv square
    1.5X10^7=(1/2)(40000)v square
    v square=750
    v=27 m/s.

    Well to be more precise its 2.1189600X10^7-5.886000X10^6=1.5303600X10^7
    1.5303600X10^7/0.15(40000)v square
    v square=765.18
    v=27.6 rounded to 28 m/s.

    So according to that method.
    I subtract Epa-Epc
    2.1X10^7 J-1.8X10^7 J
    = 3.0X10^6 J
    so 3.0X10^6 J is Ek for C?
     
    Last edited: Jun 2, 2007
  20. Jun 2, 2007 #19
    Exactly. (I think the main difference between our values is that I take g=9'8 and all numbers, sorry).
     
    Last edited: Jun 2, 2007
  21. Jun 2, 2007 #20
    its alrite and thanks
     
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