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Mass falling in Earth's Gravity help

  1. Jun 16, 2011 #1
    1. The problem statement, all variables and given/known data

    A particle falls in Earth's gravity. The force of air resistance is F = −αv v where v is the velocity, v is the speed, and α is a constant. (The direction of the force is opposite to the velocity, and the magnitude of the force is αv^2.) If the terminal speed is 20.2 m/s and the mass is 2.6 kg, determine the constant α.

    2. Relevant equations
    F=mdv/dt


    3. The attempt at a solution
    m(dv/dt)=mg-aplhav
    thus v(t)=(mg/aplha)(1-e^(-(alpha/m)t))
    so aplha if im correct should be (mg/Vterminal)=1.263, however im not sure of the units for this. My guess would be kg/s however my homework programs says that is incorrect, and also I feel I may have neglected the [/B]v[/B] and the magnitude of the opposing force causing a wrong answer.
    If someone could shed some light on this it is greatly appreciated.
     
  2. jcsd
  3. Jun 16, 2011 #2

    rock.freak667

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    From your formula it seems that you need to use

    F=-αv2 so you'd have

    ma = mg - αv2
     
  4. Jun 16, 2011 #3
    If thats the case then what does the F=-αv mean? To me, the question is pretty confusing as far as what goes where.
     
  5. Jun 16, 2011 #4

    gneill

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    At terminal velocity the velocity is constant. So the acceleration must be nil. What can you say about the forces acting in that case?
     
  6. Jun 16, 2011 #5
    it would be dv/dt=a=0 thus the net forces are balanced, I guess my confusion is what is the question asking for as the resistive force F=-αv or F=-αv^2?
     
  7. Jun 16, 2011 #6

    gneill

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    Well, I'd go with the magnitude of the resistive force and make the appropriate assumption about its direction for a body falling straight down :wink:

    I do find the notation vv a bit odd. I mean, why not v with a unit vector for direction? Maybe it's just me...:smile:
     
  8. Jun 16, 2011 #7

    Dick

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    The magnitude of F = −αv v, where the second v is a vector is -av^2. And it points in a direction opposite to the vector v if a is a positive constant.
     
  9. Jun 16, 2011 #8

    Dick

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    I think the problem statement says the magnitude is av^2. So I don't think they mean the vector v to be a unit. As you said.
     
    Last edited: Jun 16, 2011
  10. Jun 16, 2011 #9

    gneill

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    Hi Dick, That was my point; it wasn't a (magnitude)*(unit-vector) construct, but rather a magnitude*vector (vv as opposed to v2u).

    On reflection, I can see the utility of doing it this way when the magnitude you want is the square of the vector. vv rather than v2*(v/v).
     
  11. Jun 16, 2011 #10

    Dick

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    Hi gneill, yeah, I realized what you meant after I posted. Nothing wrong with the unit vector notation either. At least it makes the velocity dependence clearer than it was to kraigandrews.
     
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