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Spring Forces-Diving Board question

  1. Oct 27, 2007 #1
    Hey, I solved this problem but I need someone to double check it. Please and Thank you!

    1. The problem statement, all variables and given/known data

    A 70-kg diver on a springy diving board makes her approach. The diving board sits 1 m above the water below. As she approaches the end of the board she leaps into the air 0.5 meters. She comes down on the board with some speed causing the board to bend. Think of the diving board as a spring with a spring constant of 10,000 N/m. The board then flings her up into the air to some height.

    a) To what height above the water does the board send her?

    b) What is surprising about your result? What makes it more believable in real life?


    2. Relevant equations

    (KE+PEg+PEs)i = (KE+PEg+PEs)F

    E=(KE+PEg+PEs) = (.5)mV[tex]_{2}[/tex]+mgx+(.5)kx[tex]_{2}[/tex]

    x= E/mg

    3. The attempt at a solution

    I found the distance the board bends:

    (KE+PEg+PEs)i = (KE+PEg+PEs)F
    0+mg(h+d)+0=0+0+(.5)kd[tex]_{2}[/tex]
    (70)(9.8)(.5+d)=(.5)(10000)d[tex]_{2}[/tex]
    d[tex]_{2}[/tex]-(.1372)d-(.0686)=0
    I used the quadratic formula to get d=0.339m

    I found the energy:

    E=(KE+PEg+PEs) = (.5)mV[tex]_{2}[/tex]+mgx+(.5)kx[tex]_{2}[/tex]
    E=0+PEg+PEs)=0+(7)(9.8)(-.339)+(.5)(10000)(-.339)[tex]_{2}[/tex]
    E=(-232.554)+(547.605)=342.051

    I found the height:

    x= E/mg
    x= (342.051)/(70)(9.8)=.4986
    .4986+1=1.4986 m

    The thing here, is it realistic for her to just have a maximum height of 1.4986m? It seems like all I did was add the 0.5 m that she jumped with the height the board is above the water(1m) but as you can see, I didn't. Thus my question is if 1.4986 is my final answer, it that realistic? And I don't know how I should interpret the b part of the problem. Should my answer be coming out to be an outrageously small or big number? Or is a number around 1.5 the answer I should be getting? Please help because I'm just confusing myself.
     
  2. jcsd
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