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Gravity Problem

  1. Mar 8, 2005 #1
    Alrite, I am in gr. 11 academic physics, and have a pretty good understaning of it so far. Recently we were given an assignment and i need some clarifcation for this one question.

    The question:
    a 80kg object is free falling and when it reaches a velocity of 20m/s (down) air resistence acts on it to make it slow down at 0.2m/s/s (up)
    Find the air resistence.
    Also find its final velocity and how far it falls after 3 seconds of the force acting. How long after it was released from rest did the air resistence begin to react.

    What i need clatriy on is air resistence. Would you use Fnet=gm, or Fnet=ga to find it and do u have to use the initial velocity somehow. Or is initial velocity only there for the other parts of the question? i think my answer for the air resistence was around 1.96N.
     
  2. jcsd
  3. Mar 8, 2005 #2

    tony873004

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    I would guess that it wants the force from air resistance.
    f=ma
    f = 80 kg * 0.2 m/s/s = 16N

    How long after it was released did the air resistance begin to react. It started after it reached 20 m/s, and it accelerated at 9.8m/s/s to get to 20 m/s.

    It's final velocity after 3 seconds:
    It spent 2 seconds accelerating at 9.8 m/s/s
    and
    1 second accelerating at 9.8-0.2 m/s/s

    Your distance will be your average velocity * 3 seconds
     
  4. Mar 9, 2005 #3

    HallsofIvy

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    "What i need clatriy on is air resistence. Would you use Fnet=gm, or Fnet=ga to find it and do u have to use the initial velocity somehow. Or is initial velocity only there for the other parts of the question? i think my answer for the air resistence was around 1.96N."

    net force is neither gm not ga (ga makes no sense at all! g is itself an acceleration).

    The question seems to me to be ambiguous. Does "air resistence acts on it to make it slow down at 0.2m/s/s (up)" refer to the net acceleration or only the acceleration due to air resistance, ignoring gravity? I assumed net acceleration in what I did below. tony873004 assumed only air resistance and, upon reflection, I think he is right.

    The net force is -mg+ Fair[/b] where "Fair" is the force of air resistance. You know (I hope) that g is 9.8 m/s2 so -80(9.8)+Fair= (80)0.2 m/sec2. Fair= 80(0.2+ 9.8)= 80(10)= 800 Newtons (upward). I have no idea how you got 1.98!
    (The 16 Newtons tony873004 got was the net force- including gravity.)

    Apparently, according to this problem, there was no air resistance until the downward speed was 20 m/s. How long does it take something, accelerating at 9.8 m/s2 to reach 20 m/s?
     
  5. Mar 9, 2005 #4
    I'd skip the over-complication and use these formulas:

    [tex]F_d=C_d \frac{1}{2} \rho v^2[/tex]

    This would be the force of drag formula. I know that at terminal velocity [itex]F_d=mg[/itex], Therefore

    [tex]C_d= \frac{mg}{\frac{1}{2} \rho v^2}[/tex]

    Please correct me if I'm wrong, though, as the assignment really didn't say much about what kind of figure they were looking for, wether it was force or coefficient...
     
  6. Mar 9, 2005 #5
    Thanx for all the help guys. I did some stupid mistakes when doing the question. I tired again, and got pretty much the same as tony did, so thats good.
    thanx again
     
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