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Spring question!

  1. Feb 24, 2004 #1
    hello, if anyone can help me figure this out, it will be appreciated.
    the problem is the following:
    An 80.0-kg man jumps from a height of 2.50 m onto a platform mounted on springs. As the springs compress, he pushes the platform down a maximum distance of 0.240 m below its initial position, and then it rebounds. The platform and springs have negligible mass.
    a)What is the man's speed at the instant he depresses the platform 0.120 m?
    b)If the man just steps gently onto the platform, what maximum distance would he push it down?

    I found the potential energy when the man hits the spring by U=mgh,
    i also need the spring constant and this is where i get confused on how to get it.
    Do i use F=kx, the F being the weight???and then plug in to find w=0.5kx^2, and then finally find v??
    I also get confused as to which distances to use on the equations
    Any help would be great!
    thanks.

    [?]
     
  2. jcsd
  3. Feb 24, 2004 #2

    NateTG

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    At the point where the spring is fully compressed you've got:
    [tex]PE_{spring}=mgh[/tex]

    for the second part you've got:
    [tex]kx=mg[/tex]
    which is also relatively easy to solve.
     
  4. Feb 24, 2004 #3
    thanks for ur reply NateTG, but still I get confused on how to get the constant k, on the equations u sent, they use x as distance, which distance should i use??
     
  5. Feb 24, 2004 #4

    NateTG

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    Let's find [tex]k[/tex]
    [tex]mgh=\frac{1}{2}kx^2[/tex]
    so
    [tex]k=\frac{2mgh}{x^2}[/tex]
    where h is the height the man dropped, and x is the spring displacement.
     
  6. Feb 24, 2004 #5
    hey thanks!, i got the first part for the velocity ,but the equation that you gave me for the second part, to find the distance as the man steps on the spring kx=mg is not giving me the right answer, is there another way of finding that distance???
     
  7. Feb 24, 2004 #6

    NateTG

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    I get
    [tex]k=373332 N/m[/tex]
    so I get a displacement of [tex]2cm[/tex] what do you get?
     
  8. Feb 24, 2004 #7
    i get for the first part k=68055 N/m, which gave me a velocity
    of 6.06 m/s, which is right.
    But when i use that same k for the second part, i get 0.0115 m
    for the displacement, and it says its wrong.
     
  9. Feb 25, 2004 #8

    NateTG

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    Hmm, do you know what the correct answer is supposed to be? You would get a different equation if you try disregard friction, and solve part 2 with energy.
     
  10. Feb 25, 2004 #9
    no, i don't know the answer,i put the answer on a website and it tells me wheter i'm right or not. I still have 6 tries left :wink:
    I supposed the answer would be smaller than 0.240 m which was the compression when the man jumped, I plugged in yours, 2 cm , 0.02 m , and is also wrong.i'll keep trying with what u said about not using friction and doing it with kinetic, thanks.
     
  11. Feb 25, 2004 #10
    hey, i think they wanted more sig figs, the anwer was 0.0210 m,
    i just requested it, because i got the same as you did when i worked it out.
     
  12. Feb 25, 2004 #11
    mgy=1/2kx^2

    mgy=1/2kx^2

    is all you need for these problems

    hold on While I do the calculations
     
  13. Feb 25, 2004 #12
    Worked out

    MGY
    80*2.5*9.8=1960
    1960=1/2K*.254^2
    K=68055.55556
    Then you can F=-kx
    F=-16333.3333

    you shoudl be able to solve the rest fairly easily... if i did it right hope i helped in some way
     
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