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Acceleration and terminal speed problem

  1. Feb 14, 2008 #1
    1. The problem statement, all variables and given/known data

    A small piece of Styrofoam packing material is dropped from a height of 2.10 m above the ground. Until it reaches terminal speed, the magnitude of its acceleration is given by a = g - bv. After falling 0.400 m, the styrofoam effectively reaches its terminal speed, and then takes 4.90 s more to reach the ground.

    (a) What is the value of the constant b?
    (c) What is the acceleration when the speed is 0.150 m/s?

    2. Relevant equations

    a = g-bv
    - (a - g)/v = b
    b = mg/vt (Don't know mass or terminal velocity.

    3. The attempt at a solution

    I know that after the Syrofoam reaches it's terminal speed acceleration goes to 0 and the velocity is constant. I don't know the acceleration or really the velocity to solve for b. I think c can be solved once a is figured out, but I'm not sure how to get there.
     
  2. jcsd
  3. Feb 14, 2008 #2
    I actually just solved this one. My last problem is this:

    Calculate the force required to pull a copper ball of radius 1.50 cm upward through a fluid at the constant speed 9.00 cm/s. Take the drag force to be proportional to the speed, with proportionality constant 0.950 kg/s. Ignore the buoyant force.

    So I would use R = -bv to calulate the force. B = .950 kg/s. V = .090 m/s.

    I need to use F = mg in there right to sum the forces. so F = -bv - mg? How do I get the mass? Am I on the right track? Thank You.
     
  4. Feb 14, 2008 #3

    mgb_phys

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    They said ignore the buoyant force so you don't need to know the mass.
    It's just a "pulling object against friction" question with the complication that the friction depends on speed.
     
  5. Feb 14, 2008 #4
    So the radius has nothing to do with the problem? Is it just R = -bv?
     
  6. Feb 14, 2008 #5

    mgb_phys

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    Drag force is F = - b v for low velocities in a fluid.
    The constant b is proprtional to radius,
    b = 6 pi eta r where eta is the viscosity in kg / m / s, r is the radius, so the units of b should be kg/s as given.
     
  7. Feb 14, 2008 #6
    Well F = -bv is equal to (.950k/s * .09 m/s) = .0855, which is not correct. What is missing? Where does gravity plan in? It's pulling down as we are lifiting it up.
     
  8. Feb 14, 2008 #7

    mgb_phys

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    I would have said that 'ignore bouyant force' also means ignore gravity, it could be being pulled through molten copper in which case the weigth and bouyant force cancel.
    on the other hand it could mean you should include the weight but ignore any upthrust from the fluid ( assume it is in helium or some very light gass with a very large viscous drag!!!)
    Try adding the weight and see if that's the right answer ( density of copper = 9g/cc)
     
  9. Feb 14, 2008 #8
    Nope, I am looking for an answer that is around 3N. I am not where close.
     
  10. Feb 14, 2008 #9
    Any other ideas?
     
  11. Feb 14, 2008 #10

    mgb_phys

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    No, I get the weight to be 1.25N and the viscous drag to be 0.09N like you do - sorry
     
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