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How far can a 55kg person fall and break and leg?

  1. May 8, 2009 #1
    Hi, having a problem with this question. I thought it was as simple as the conservation of energy, but I guess there is something I'm missing. The program we're using isn't accepting my answer, even when I use two significant digits.

    Thanks for the help,

    Cohort

    1. The problem statement, all variables and given/known data
    From approximately what maximum height could a 55 kg person jump and land rigidly upright on both feet without breaking his legs?

    The energy a human leg bone can absorb is 200 J.
    The person is 55 kg.

    2. Relevant equations

    mgy = 1/2mv^2


    3. The attempt at a solution

    (55 kg) (9.8 m/s^2) y = 200 J
    y = 200 J/ 539
    y = .37 m
     
    Last edited: May 8, 2009
  2. jcsd
  3. May 8, 2009 #2

    nvn

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    cohort: Generally always maintain four significant digits throughout all your intermediate calculations, then round only the final answer to three significant digits, unless the first significant digit of the final answer is 1, in which case round the final answer to four significant digits. Have you tried this yet? Note also that numbers less than 1 should always begin with a zero before the decimal point. Omitting the zero violates ISO 31-0. How many tries do you have remaining? Your answer looks correct to me, so far, unless I am missing something.
     
  4. May 8, 2009 #3
    I've tried so many times I might not get credit for it, but I don't want to choose "show answer" in case I can still get credit for it.

    55 kg * 9.8 m/s^2 = exactly 539 N. 200 J / 539 N = .3710575139 They want two significant digits so I put in 3.7*10^-1, but it still said Try Again. This is frustrating.

    Should I just hit Show Answer, or is there some other way of entering the height I haven't tried yet?

    Thanks for the help.

    Edit: The zero thing didn't work either.

    By the way, I'm using mastering physics.
     
  5. May 8, 2009 #4

    nvn

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    cohort: No, don't hit "show answer" yet. Let us think it over awhile. When is the deadline? Wait until just before the deadline. Have you tried entering 0.37 and 0.371 yet? Regardless, don't hit "show answer" yet if you don't have a deadline yet.
     
  6. May 8, 2009 #5
    Midnight. Its one of the last two problems I have left.

    I tried 37 * 10^-2 also, and it didn't work.
     
  7. May 8, 2009 #6

    nvn

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    cohort: Have you tried entering 0.37 and 0.371 yet, without scientific notation? Secondly, are units also required by the program when inputting an answer, such as 0.37 m or 0.371 m? How many hours from now is midnight?
     
  8. May 8, 2009 #7
    I tried both of those. Neither worked. Units are automatically included in the answer.

    It is 10:29 where I am.

    I'm thinking I'll just e-mail my teacher and tell him the program is wrong. It has to be!
     
  9. May 8, 2009 #8

    nvn

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    Wait an hour before doing anything. Let us think it over awhile.
     
  10. May 8, 2009 #9

    nvn

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    Hint: How many legs does a typical human have?
     
  11. May 8, 2009 #10
    2. Oh, wait so its twice as high!

    Haha. Thank you so much. It was the little details that got me on that one. Just one more problem to go and I'm done!
     
  12. May 8, 2009 #11

    nvn

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    Did it give you credit? Or did it say you had too many tries?
     
  13. May 8, 2009 #12
    I got credit. A little under half, but its better than nothing. Thanks again!

    If you could, I'm stuck on this last question too. Its finding the power it takes to move a 2300 N elevator at a constant velocity of 7.0 m/s.

    Work = force x distance. I figure 2300 N over 7 m = 16100 J.

    Power = work over time, so 16100/ 1 second = 16100 kw.

    I sounds right to me, but am I missing something?

    Edit: BrB, drive home.
     
    Last edited: May 8, 2009
  14. May 8, 2009 #13

    nvn

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    If this helps, the units on your final answer are wrong.
     
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