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Quick static fluids question

  1. Dec 8, 2007 #1
    1. The problem statement, all variables and given/known data

    A small sphere 0.75 times as dense as water is dropped from a height of 11 m above the surface of a smooth lake. Determine the maximum depth to which the sphere will sink. Neglect any energy transferred to the water during impact and sinking.

    for clarity's sake, i'm letting v = volume, and v = velocity

    2. Relevant equations

    vf^2 = Vi^2 + 2ad, Fb = pgv, K = 1/2mv^2, Work(non conservative) = change in mechanical energy


    3. The attempt at a solution

    the first thing i did was basic kinematics to get a velocity of 14.68 m/s upon hitting the water.
    next i set Fb equal to nonconservative work and i got pgv(d) = Kf -Ki...
    simplified down i got it to pvg(d) = 1/2mv^2
    = pvg(d) = 1/2pvv^2

    at this point i cancled out the volumes and got a final equation of
    1000*9.8*(d) = .5*750*(14.68^2)

    this got me a final answer of 8.25 meters which seems to be too large, and in fact it was wrong.
    can anyone find where i'm going wrong?
     
  2. jcsd
  3. Dec 8, 2007 #2

    Doc Al

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    Staff: Mentor

    Gravity still acts on the object when it's under water.
     
  4. Dec 8, 2007 #3
    oh i see, but how do i take that into account? the buoyant force still is non conservative right?
     
  5. Dec 8, 2007 #4

    Doc Al

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    Yes, buoyant force is the only non-conservative force acting (in this simplified problem).

    You want the super-duper easy way to solve this? Measured from the final position of the object (distance d under the water), what's the total mechanical energy of the object just before it's dropped? After it reaches the final position? So what must the non-conservative work equal?
     
  6. Dec 8, 2007 #5
    or can i make an Fnet equation

    like, mg - Fb = ma

    p(ball)vg - p(water)vg = p(ball)*v*a

    volumes cancle, so => 750(9.8) -1000(9.8) = 750*a
    this yielding an acceleration of 3.267 m/s^2

    would that be a correct approach, or is what i just laid out above incorrect?
     
  7. Dec 8, 2007 #6

    Doc Al

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    Nothing wrong with that approach. Go for it.
     
  8. Dec 8, 2007 #7
    ok so since theres potential at the top and nothing(?) at the end, can i say

    p(water)vg(d) = p(ball)vgh

    1000(d) = 750(11)

    d = .12 meters????

    i really appreciate the help
     
  9. Dec 8, 2007 #8

    Doc Al

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    That "h" has to be the total change in height, not just the height above the water surface.
     
  10. Dec 8, 2007 #9
    ok that makes sense, but won't that leave me with two unknowns?
     
  11. Dec 8, 2007 #10
    or can i make h = 11+d?

    but when i did that i got a final d of 33 meters. this seems way too large to me
     
  12. Dec 8, 2007 #11

    Doc Al

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    Looks OK to me. Solve it the other way and compare.
     
  13. Dec 8, 2007 #12
    i just don't understand how it can sink that far...

    btw thanks for all the help, i really understand how to solve this type of question now, i appreciate it, thanks!
     
  14. Dec 8, 2007 #13

    Doc Al

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    That's because you have some common sense. :smile:
    The problem is the bogus assumption: "Neglect any energy transferred to the water during impact and sinking." That's a huge effect to ignore!
     
  15. Dec 15, 2007 #14
    im still confused, my height above water was 12 instead of 11, and i tried what he did, with 1000(d)=750(12+d) because h=12+d..... i got my h to be 48 and d to be 36... i tried both those answers and neither worked. his was the correct answer and im pretty sure what i did is what he did... how did he get 33 for his final answer?

    nvmmmm i used the wrong coefficients!
     
    Last edited: Dec 15, 2007
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