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Ball falling through a liquid

  1. Nov 5, 2008 #1
    3. The attempt at a solution

    I think the first part is just 9.8 but I have no idea on the rest of the problem (this is due tomorrow need help)
     

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  3. Nov 5, 2008 #2

    Moonbear

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    You need to show some of your own effort if you wish to receive help.

    What have you tried so far? What equations or concepts do you think might be related to this problem?
     
  4. Nov 5, 2008 #3
    I would think that at the begging only gravity would act on it so it would be 9.8

    Then for part two all I can get is that F=kv and it is given that F=-bmv so I think that k=-bm then I have an equation here that says that terminal velocity will equal Vt=mg/k, pluging k into the equation will get -g/b (is that right)

    For the final part i really don't know where to start (i am in calculus this year).
     
  5. Nov 5, 2008 #4

    PhanthomJay

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    Your answers to a and b are correct, but I'm not sure if you're just blindly pulling equations out of somewhere without understanding their derivations. The differential equation is difficult to solve, but it's not that hard to write it down, because it is just Newton's 2nd law for constant mass: [tex]F_{net} = ma[/tex], or in its calculus based differential form, [tex]F_{net} = m(dv/dt)[/tex]. So what is [tex]F_{net}[/tex] on the ball?
     
  6. Nov 5, 2008 #5
    Can you have F=-bmv which means that acceleration equals -bv? then i'm not sure about the calculus on how to get time factored into the equation.
     
  7. Nov 5, 2008 #6

    djeitnstine

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    [tex]\sum \vec{F} = m \vec{a}[/tex]

    [tex]-bv + mg = m\vec{a}[/tex]

    [tex]mg - bv = m \frac{d{v}}{dt}[/tex]

    [tex]g - \frac{bv}{m} = \frac{dv}{dt}[/tex]

    [tex] g = \frac{dv}{dt} + \frac{bv}{m}[/tex]

    Here you get a differential eq. Which turns out to be

    At terminal velocity a = 0

    [tex] V_{terminal}= \frac{mg}{b}(1-e^{\frac{-bt}{m}})[/tex]


    There is everything you will need to solve that problem.
     
  8. Nov 6, 2008 #7

    PhanthomJay

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    you are not applying Newton 2 correctly. You've got to look at all the forces acting on the ball, not just the resistive force. The resistive fluid drag force on the ball is given as 'bmv' acting up. There's another force on the ball acting down....please identify it. Then determine the net force which will be the algebraic sum of thise 2 forces, and set it equal to 'ma'. Your correct answers to parts a and b will come directly from that equation, using the known conditions that v=0 when the ball is first dropped into the water, and a = 0 at terminal velocity.
     
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