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Newtons first and second laws

  1. Sep 19, 2008 #1
    1. A 3600 jet touches down at 250 on the deck of an aircraft carrier and immediately deploys a parachute to slow itself down. If the plane comes to a stop in 180 , what is the average force of air on the parachute? Assume the parachute provides essentially all the stopping force.



    2. Fnet= mass*acceleration, a=v/t, v=p/t



    3. M=3600kg
    v=250km/h
    p=180


    v=p/t 180m/25000m/h convert the kilometers to meters t=7.2e^-4*3600
    t=2.592



    250000m(1/3600sec)= 69.4 m/s



    fnet=ma

    m=3600


    fnet=3600kg(69.4(m/s)/(2.592 s)= 96450N


    i tried doing it but i keep on getting it wrong.
     
  2. jcsd
  3. Sep 20, 2008 #2

    Kurdt

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  4. Sep 20, 2008 #3
    ]1. A 3600kg jet touches down at 250km/h on the deck of an aircraft carrier and immediately deploys a parachute to slow itself down. If the plane comes to a stop in 180m , what is the average force of air on the parachute? Assume the parachute provides essentially all the stopping force.


    I was thinking about using v2=v02+2a deltat
     
  5. Sep 20, 2008 #4
    am i on the right track if i do that
     
  6. Sep 20, 2008 #5
    If you're asking whether or not using the equation v squared = v0 squared + 2as would help you, then yes you would be on the right track.

    Another way to solve this would be by using the impulse momentum theorem Ft = change in momentum.

    Hope this helps!
     
  7. Sep 20, 2008 #6

    Kurdt

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    Yes you can use that equation to find acceleration.
     
  8. Sep 20, 2008 #7

    cepheid

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    Wouldn't it make more sense to use the work-energy theorem? I mean, you're given the distance over which the force acts, not the time interval.
     
  9. Sep 20, 2008 #8
    Now that I think about it yes :)

    Just goes to show that there are many different ways to solve a given problem.
     
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