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Kinetic Energy - Major Help Needed

  1. Jan 20, 2005 #1
    Kinetic Energy -- Major Help Needed

    I'm not even looking for answers ... just for somebody who is really willing to help me out here because I definitely do not understand ANY of this stuff that we're doing right now. Any body, please?
  2. jcsd
  3. Jan 20, 2005 #2
    What do you need help on?
  4. Jan 20, 2005 #3
    Heh .. this is just the first problem of my homework that I entirely DO NOT understand ...

    In 1994, Leroy Burrell of the US set what was then a new world record for the me's 100 m run. He ran the 1.00 X 10^2 m distance in 9.85s. Calculate burrelll's kinetic energy, assuming that he ran w/ a constant speed equal to his average speed. Assume his mass was 75.0 kg!

    So when I work this all out. .. I got this number such as 3865.08 J ..... but that isn't even possible -- I'm so lost :(
  5. Jan 20, 2005 #4


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    Do you know [tex]KE=(1/2)mv^2[/tex]?
  6. Jan 20, 2005 #5
    well, I guess that maybe it could be right ... it is just .. why do I have to answer it w/ it being 3.87 X 10 ^3? that kind of confused me ...
  7. Jan 20, 2005 #6


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    Actually I just did the problem and I believe your answer is right. Don't take my word for it though.
  8. Jan 20, 2005 #7
    Maybe I just need to learn how to become more confident with this stuff ... but I just can't visualize it very well ..... any suggestions as to how I can start to really visualize this stuff more? One final question for the time being ... how do I convert from MJ to J?
  9. Jan 20, 2005 #8


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    You did this right?




    [tex]Ke=3865.082839 J[/tex]

    Seems right to me
  10. Jan 20, 2005 #9
    alright now ... say i'm trying to find the kinetic energy and i have an average speed .... ?
  11. Jan 20, 2005 #10


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    1 MJ= 1 000 000 J

    Maybe you mean an accelerated speed, where you can get the average acceleration:

    [tex]a_{average}=\frac{v_2-v_1}{t_s} [/tex]

    [tex]v_2[/tex] is the final speed
    [tex]v_1[/tex] is the initial speed.
    t is time in seconds
  12. Jan 20, 2005 #11
    Thank you :)
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