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Kinetic energy

  1. Dec 3, 2003 #1
    Ok, I have tried to rationalize this problem, but I am stuck, I don't even know how to start! I know it has to do with kinetic energy though.
    When a truck goes from rest to 30 miles per hour, it uses one ounce of fuel. HOw many ounces of fuel does it use when it goes from 30 miles per hour to sixty miles per hour?
     
  2. jcsd
  3. Dec 3, 2003 #2

    krab

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    Fuel quantity is a measure of energy quantity. Do you know how energy depends on speed?
     
  4. Dec 3, 2003 #3
    the more speed you need, the more fuel you need, so the more energy you need?
     
  5. Dec 3, 2003 #4

    Doc Al

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    You said it yourself in the title of this thread. Kinetic Energy! How does that depend on speed?
     
  6. Dec 3, 2003 #5
    If the truck were in a frictionless world, and the engine of the truck had the same HP at all rpm's, then the energy needed to accelerate 30mp/h would be the same at all speeds. hmmmmm... The rpm's would increase as the truck gained speed, so the amount of fuel used would be greater each progressing second. The rpm increase would mean that more fuel is burning and so more energy is being produced. But in the real world the friction would also increase and would limit the power of the engine. So to answer your question, if you are considering that there would be no friction(or loss of energy due to heat, sound, light, electricity, or magnitism), then the amount of fuel needed to accelerate from 30mp/h, to 60mp/h, would be 1 ounce of fuel.

    It would be much harder to solve this and include all the other factors.
     
  7. Dec 3, 2003 #6

    Doc Al

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    I believe you are overcomplicating what was probably meant as a simple problem, as well as missing a key point. Look at it this way: The car is device for transforming chemical energy into kinetic energy (with a certain efficiency). We are told that 1 ounce of fuel can be transformed into the KE associated with 30 mph (KE30). So what's the additional KE needed to get to 60 mph (KE60)? And how much fuel does that represent?
     
  8. Dec 3, 2003 #7
    I believe you are overcomplicating what was probably meant as a simple problem, as well as missing a key point. Look at it this way: The car is device for transforming chemical energy into kinetic energy (with a certain efficiency). We are told that 1 ounce of fuel can be transformed into the KE associated with 30 mph (KE30). So what's the additional KE needed to get to 60 mph (KE60)? And how much fuel does that represent?

    I believe I may have over simplified the question, so i don't know how you can say that. I answered the question in the most accurate way and avoided making any assumptions on the efficiency of the truck. If the truck were for real then to go from 30mp/h to 60mp/h would be > 1 ounce of fuel.
     
  9. Dec 4, 2003 #8
    Is this all the info? If so, then assume no friction, and the answer is 1 ounce of fuel.
     
  10. Dec 4, 2003 #9

    chroot

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    No. Kinetic energy goes as the square of the velocity. A truck moving at 60 mph has four times as much kinetic energy as a truck moving at 30 mph. If the truck uses one ounce of fuel to accelerate to 30 mph, it must therefore use three more ounces of fuel to accelerate from 30 to 60 mph.

    - Warren
     
  11. Dec 4, 2003 #10
    I don't see why this would be so. 60²=3600>120=30x4 so what you said makes no sence,cause the numbers don't add up.
     
  12. Dec 4, 2003 #11

    Doc Al

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    The KE at 60 mph is 4x the KE at 30 mph. If one ounce of fuel provides the energy to get to 30 mph, then you need three more ounces to get to 60 mph.

    (Compare 602 to 302, not 30)
     
  13. Dec 4, 2003 #12

    chroot

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    The kinetic energy of an object moving with speed [itex]v[/itex] is

    [tex]K = \frac{1}{2} m v^2[/tex]

    The kinetic energy of an object moving at 60 mph is

    [tex]K = \frac{1}{2} m\ 60^2[/tex]

    The kinetic energy of an object moving at 30 mph is

    [tex]K = \frac{1}{2} m\ 30^2[/tex]

    Thus the ratio of the kinetic energy of the 60 mph object to the 30 mph object is

    [tex]
    &\frac{(1/2) m\ 60^2}{(1/2) m\ 30^2} = \frac{60^2}{30^2}
    = 4
    [/tex]

    Get it? Double the velocity, quadruple the kinetic energy.

    - Warren
     
  14. Dec 5, 2003 #13
    Thank you!
     
  15. Dec 8, 2003 #14
    I would agree with chroot.
    If increasingly more energy was not required for each additional velocity increase, than we could easily construct vehicles to travel the speed of light!
     
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