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Homework Help: Kinetic Energy/Power Question

  1. Feb 25, 2010 #1
    I just have some questions about kinetic energy.

    1. If you are going up an incline on a roller coaster and you are trying to calculate the KE lost; is it possible to use this to find the velocity at the top, or does this only apply to downwards motion?

    2. Can you apply the power formula to a roller coaster going down an incline, or is this only useful for when a motor is in charge of the energy outlay rather than gravity?

    3. Can you use GPE and KE, even if you are on an incline/slope?
     
  2. jcsd
  3. Feb 25, 2010 #2

    tiny-tim

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    Hi Procrastinate! :smile:
    Yes, KE + PE = constant works in all circumstances (assuming, of course, no energy loss to friction etc). :smile:

    Incidentally, potential energy (PE) is just another name for (minus) work done by a conservative force (and gravity is conservative). :wink:
    Sorry, what's the "power formula" ? :confused:
     
  4. Feb 25, 2010 #3
    P = E/t - is this only used when a motor is involved, rather than being controlled by gravity.


    So, back on Kinetic energy. If say for example, a 20kg mass was moving up an incline from a height of 3m to a height of 10m and I wanted to find the final velocity at the top? I would do:

    GPE_top - GPE_bottom = KE lost=1372

    1/2xmv2 = KE lost
    1372=.5mv2
    11.7=velocity
     
  5. Feb 25, 2010 #4
    well in case of gravity power isnt dissipated but changed into another form
    but i think u can stil apply that formula(actually more proper would be to apply f.v though it is very much derived from e/t) and u will know the ke that converted into potential

    well not sure about ur calculation but the method for the example is right
     
  6. Feb 25, 2010 #5

    tiny-tim

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    ah, that's the definition of power … power = energy (or work) per time.

    you can only really use power when for some reason you know exactly what it is … eg your example of a motor, with a given power.

    (and I personally have never seen an exam question involving power, so you can forget it for solving rollercoasters etc :wink:)
    That's fine, except the last two lines …

    1372 is the difference in KE …

    to find the final velocity, start with the initial velocity, and subtract 1372 from its KE. :smile:
     
  7. Feb 25, 2010 #6
    Ok, thank you.

    Just to check, can you use GPE and KE on a spiral that is going down? Or is it better to use circular motion formulas? When i am finding acceleration centripetal, I get confused as to whether to use initial velocity first going around the curve, or final velocity going around the curve.
     
  8. Feb 26, 2010 #7

    tiny-tim

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    Hi Procrastinate! :smile:

    (just got up :zzz: …)
    GPE and KE are almost always better, because they involve v instead of a …

    in other words, they automatically do the job for you of performing a possibly difficult integration!! :biggrin:
    I'm not sure what you mean :confused:

    centripetal acceleration is always v2/r, and on (for example) a rollercoaster, you can always find v using KE + PE = constant.
     
  9. Feb 26, 2010 #8

    When I am substituting v in the equation to find centripetal acceleration, and this is on a spiral that is going down, the velocity at the top of the curve will be different from the velocity at the bottom of the curve. What do I do then?
     
  10. Feb 26, 2010 #9

    tiny-tim

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    (btw, do you mean a helix?)

    The centripetal acceleration will change … is that what's confusng you? :confused:

    it will always be v2/r.

    (but why are you trying to find the centripetal acceleration anyway?)
     
  11. Feb 26, 2010 #10
    It looks a bit like this: I posted it here (I wanted to delete it but I can't): https://www.physicsforums.com/showthread.php?p=2599095#post2599095 [Broken]



    I need to compare the centripetal acceleration on my X and Z acceleration graphs from the accelerometer for the primary data that was collected.
     
    Last edited by a moderator: May 4, 2017
  12. Feb 26, 2010 #11

    tiny-tim

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    hmm … it doesn't look like there's any way to calculate r (the radius of curvature of the track) … is there?

    (also, it looks to me as if the centripetal acceleration will not be horizontal)

    If not, you'll have to get the centripetal acceleration from the accelerometer (and make allowance for g, if it's not horizontal)
     
  13. Feb 26, 2010 #12
    If I do use my acceleration graphs, which axis should I use? The Z acceleration (I got it mixed up with X before, which actually wasn't provided) or the Y acceleration?

    I found the radius of the track from Google Earth, even if it is not 100% accurate.
     
  14. Feb 26, 2010 #13

    tiny-tim

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    Sorry, I've no idea what you're talking about. :redface:

    What are these acceleration graphs? :confused:
     
  15. Feb 27, 2010 #14
    Actually, I get it now, so it doesn't matter.

    However, is it better to leave calculating centripetal motion out at the curves I mentioned in the diagram since they are not completely horizontal?
     
  16. Feb 27, 2010 #15

    tiny-tim

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    You can always use centripetal acceleration, but if the bend is on a slope, the centripetal acceleration will along that slope also. :smile:
     
  17. Feb 27, 2010 #16
    Do you mean change along the slope?


    Thank you for your help by the way.
     
  18. Feb 27, 2010 #17

    tiny-tim

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    oops!

    oops! i left out the word "be" :rolleyes:

    … the centripetal acceleration will be along that slope (and not horizontal or vertical) :smile:
     
  19. Feb 27, 2010 #18
    Re: oops!

    Thank you very much.
     
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