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Diving into water: avg F of water

  1. Apr 2, 2012 #1
    A 53.0-kg person jumps from rest off a 12.0-m-high tower straight down into the water. Neglect air resistance. She comes to rest 3.00 m under the surface of the water. Determine the magnitude of the average force that the water exerts on the diver. This force is nonconservative.
    A. 2100 N B. 7800 N C. 6200 N D. 520 N E. 2600 N

    height=12 m
    depth= 3m
    mass= 53kg

    Wgrav= mg (delta h)
    = 5194 N (should I put h or h+d??? I'm using h+d)
    =7791 J

    Actually, I am pretty lost and could use a little guidance. Thank you
     
  2. jcsd
  3. Apr 2, 2012 #2
    The kinetic energy when the diver hits the water is equal to the kinetic energy the water exerts, right? Because of cons of E? So if I can find the diver's E when she has gone 12 m, I will know the avg F?
     
  4. Apr 2, 2012 #3
    OK, she hits water level at 117.6 m/s.
    KE = 0.5 x 53 x (117.6^2) = 366488.64

    delta KE = Fd
    366488.64 = Fx3
    366488.64/3=F
    122162.9 =F
    ???

    That is not an option but it seem right to me. :(
     
    Last edited: Apr 2, 2012
  5. Apr 2, 2012 #4
    I do not believe you need to introduce velocity into your thought process.
    [tex]U_1+K_1=U_2+K_2+J[/tex]
    where J is the amount lost due to your nonconservative forces (a positive quantity). Take point 1 to be the moment before she jumps and point 2 to be her resting position underwater. At points 1 and 2, she is not moving.
    [tex]U_1=U_2+J[/tex]
    We define the coordinates such that her resting position is at zero height.
    [tex]U_1=J[/tex]
    where U_1 is the gravitational potential energy computed using a coordinate whose origin is her resting position with the positive h-axis being a straight shot from point 1 to point 2. Can you take it from here? Use the definition of energy given a force and displacement.
     
    Last edited: Apr 2, 2012
  6. Apr 2, 2012 #5
    I've never seen that equation before and I don't understand what it is. Why doesn't my solution work? I found KE? Is that equation the only way to find the solution?
     
  7. Apr 2, 2012 #6
    You've never seen the conservation of energy equation and you're learning about energy?! Let's start off with
    [tex]U_1+K_1=U_2+K_2[/tex]
    This holds for any particle's energy if it isn't losing energy somehow (friction, water absorbing energy, etc.) It states that the gravitational potential energy and the kinetic energy at one point must equal the same at another point. These points are denoted "1" and "2". We can substitute the actual ways to compute these quantities, which I'm sure you've seen already:
    [tex]mgh_1 +\frac{1}{2}mv_1^2=mgh_2 +\frac{1}{2}mv_2^2[/tex]
    We can think about this as saying an object slows down as it rolls up a hill (h_2 > h_1 means v_2 < v_1 or else the quantities will not equal) or an object is speeding up while rolling down a hill, etc.


    Ok, so the next leap is if you have energy losses like in real life. Every time a ball bounces, the floor absorbs some energy and it keeps reaching lower and lower highs. When you roll a ball on a floor, it will come to rest even on a level field, etc. What we say then is the initial kinetic energy and gravitational potential energy at point 1 is equal to that of point 2 plus the losses. Think about
    [tex]mgh_1 +\frac{1}{2}mv_1^2=mgh_2 +\frac{1}{2}mv_2^2[/tex]
    If an object is losing energy as it goes from point 1 to point 2, side 2 will actually be less than side 1. They are no longer equal:
    [tex]mgh_1 +\frac{1}{2}mv_1^2>mgh_2 +\frac{1}{2}mv_2^2[/tex]
    So we reinstate the equality by adding some positive number that I called J, which represents all the lost energy during the movement from 1 to 2:
    [tex]mgh_1 +\frac{1}{2}mv_1^2=mgh_2 +\frac{1}{2}mv_2^2 + J[/tex]

    The water is, in your problem, solely responsible for any imbalance in the equation. So to find how much work the water force did, find J. Then use J to find that force in the equation for work given a force and displacement.

    As for your question about kinetic energy, the reason you cannot use it solely is it ignores the fact that you are still losing potential energy as you go from 12meters in your fall to 15 meters. So your answer will be incorrect slightly (and that incorrect answer is one of the choices, by the way). So even if you find the kinetic energy right at 12 meters, you will still need to invoke the use of these equations to find conservation of energy. Things will only be messier 1.) since your kinetic energies are not zero for point 1 (since you will take point 1 to be right above the water) and and 2.) since you had to do some kinematics to find the velocity above the water (or some extra energy computation). It is worth noting, too, that she is not traveling at 117.6m/s at impact.
     
    Last edited: Apr 2, 2012
  8. Apr 2, 2012 #7
    I can't read any of your equations. Anyway, I realized that I calculated v incorrectly. How did I get that number? I have no idea. v = 15.33 and now that I plug that into my work, my solution is correct. Still have no idea what you're talking about but this way works. Behold my works ye mighty!!!
     
  9. Apr 2, 2012 #8
    Your answer is not correct. You are ignoring the extra energy gained as she goes from 12 meters to 15 meters.
     
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