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Work Done by a Landing Mat

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  1. Feb 26, 2016 #1
    1. The problem statement, all variables and given/known data
    Pole vaulter Yelena Isinbayeva (64.1 kg) is at a point in her vault where the pole is maximally deflected (Δx = 1.63 m). The pole has a bending stiffness of 1091 N/m and behaves according to Hooke's Law. At this point in the vault, her vertical velocity is 3.25 m/s and she is 2.48 m above the ground.

    Assume the vaulter lands on a mat whose top surface is 1 m above the ground and that the mat deforms by 0.5 m in bringing the vaulter to a complete stop. How much work was done on the vaulter by the mat?

    2. Relevant equations
    P=W/t
    work done = change in energy
    Hooke's Law: Fs=kΔx
    Ws=(1/2)kΔx2
    KE=(1/2)mv2
    PEg=mgh
    SE=(1/2)kΔx2
    Total energy = KE+PE+SE
    W=FΔd

    I have already calculated the following from previous questions
    Work done on the pole = 1449.34J
    SE stored in the pole = 1449.34J
    Force applied to the pole = 1778.33N
    KE in vertical direction = 338.53J
    PEg = 1559.48J
    Peak height of her centre of mass = 5.32m
    Work done by gravity from 2.48m above ground to peak height = -1785.80J
    Time it took to fall from peak height to mat = 0.94s
    The vertical velocity she first contacted the mat at = -9.21m/s

    3. The attempt at a solution
    My attempts are all giving me the same number so I'm pretty sure I'm thinking about it or interpreting it in an incorrect way.

    I tried
    W=FΔd
    W=[(64.1)(9.81)](1+0.5)
    W=943.23

    PEg=mgh
    PEg=(64.)(9.81)(1.5)
    PEg=-943.23

    Tried it negative the second time just to see...
     
    Last edited: Feb 26, 2016
  2. jcsd
  3. Feb 26, 2016 #2

    haruspex

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    In all those numbers, which is your final answer?
    You do not need to find the height of the vault, or the speed on hitting the mat. Just consider the total energy at the instant described and the total energy at the end. Where has the difference gone?
    I do notice what looks like a distance of 1.5m in your calculation. Where is that coming from?
     
  4. Feb 26, 2016 #3
    My final answer was 943.23J.

    The 1.5m distance is the mat's distance from the ground + how far the mat sinks down (0.5m)
     
  5. Feb 26, 2016 #4
    Thinking about what you said about total energy I thought maybe this might work:
    W=KEf-KEi
    W=(1/2)mv2-(1/2)mv2
    W=(1/2)(64.1)(0)2-(1/2)(64.1)(-9.21)2
    W=0-2718.61
    W=-2718.61J

    This turned out to also be incorrect.
     
  6. Feb 26, 2016 #5

    haruspex

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    The surface of the mat starts 1m off the ground. When depressed .5m by the athlete, how far is it off the ground?
     
  7. Feb 26, 2016 #6
    Ohhhhhhh my goodness 0.5m! Can't believe I overlooked that, I even have it drawn out on my paper... :oops:

    W=FΔd
    W=[(64.1)(9.81)](0.5)
    W=314.41J

    This is not correct when I put it into my assignment either

    I don't understand. I thought that the work required to stop a falling object was the KE, which I tried above and was not correct.
     
  8. Feb 26, 2016 #7

    haruspex

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    Yes, you have to take all the energy into account. All that was wrong with your previous attempt (I think) is you had the wrong final height from the ground.
    Go back to your first post where you calculated all the energies at the given initial instant. Figure out the final total energy (i.e. at a height of 0.5m), and calculate the lost energy.
     
  9. Feb 26, 2016 #8
    Okay,

    TE=PEg+SE+KE
    TE=1559.48+1449.34+338.53
    TE=3008.82

    But make it -3008.82 because it is opposite the direction Yelena is travelling.
    My online assignment told me this is correct. Thank you for your help! :)
     
  10. Feb 26, 2016 #9

    haruspex

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    Ok!
     
  11. Feb 29, 2016 #10
    How did you get 3008.82 for total energy when I did it I got 3347.35
     
  12. Feb 29, 2016 #11

    haruspex

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    I assume x2017's last post left out part of the working. As noted in the thread, the remaining PE (0.5m above ground) needs to be subtracted from the energy absorbed. That doesn't quite seem to fix up the discrepancy, but it's close. If you are still bothered, please post your working.
     
  13. Feb 29, 2016 #12
    The above image shows Yelena Isinbayeva (60.1 kg) at a point in her vault where the pole is maximally deflected (Δx = 1.60 m). The pole has a bending stiffness of 1000 N/m and behaves according to Hooke's Law. At this point in the vault, her vertical velocity is 4.18 m/s and she is 2.41 m above the ground.

    Assume the vaulter lands on a mat whose top surface is 1 m above the ground and that the mat deforms by 0.5 m in bringing the vaulter to a complete stop. How much work was done on the vaulter by the mat?
    I had the same question but different numbers.
    For PE I got 1420.9
    KE=525.04
    SE= 1280
    so I did the same thing for TE=1420.9+1280+525.04=3226
    It wasn't the right answer so I don't know what I'm doing wrong.
     
  14. Feb 29, 2016 #13

    haruspex

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    At what height does she come to rest?
     
  15. Mar 1, 2016 #14
    Her peak height is 5.47 m and she lands on the mat and is depressed by 0.5m
     
  16. Mar 1, 2016 #15

    haruspex

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    In this way of solving the problem (comparing total energy at the described instant with total energy at the end) the peak height is irrelevant.
    I asked what height she is off the ground at the end.
     
  17. Mar 1, 2016 #16
    the mat is 1m above the ground and she depresses it to 0.5m so I guess she would be 0.5 m off the ground, its the same question as the above one but with different numbers.
     
  18. Mar 1, 2016 #17

    haruspex

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    yes. Did you take that into account in your energy calculation?
     
  19. Mar 1, 2016 #18
    I'm not sure how you would take in into account, how would you go about doing that?
     
  20. Mar 1, 2016 #19

    haruspex

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    How did you calculate the PE in post #12?
     
  21. Mar 1, 2016 #20
    Never mind I got it, thanks for the help.
     
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