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Air Resistance

  1. Nov 9, 2014 #1

    SCA

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    1. A parachutist of mass 62 kg is falling at 55 km/h when her parachute open. She then falls 24 m in the next 2.0 seconds. Find the force of air resistance acting on the parachute during those 2.0 seconds.


    2. Could I just find the force of friction?


    3. I found out what the force would be if not including air resistance 607.6N but I do not know how to find the force of air resistance
     
  2. jcsd
  3. Nov 9, 2014 #2

    PeroK

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    If someone falls 24m in 2 secs, what is the most obvious thing you can calculate from that?
     
  4. Nov 9, 2014 #3

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    Speed is 12m per second
     
  5. Nov 9, 2014 #4

    PeroK

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    Is that a constant speed?
     
  6. Nov 9, 2014 #5

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    Yes
     
  7. Nov 9, 2014 #6

    PeroK

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    Are you sure about that?
     
  8. Nov 9, 2014 #7

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    Nope, I have no idea where to even start but I did try using:
    Fnet=ma
    which is Fnet=3410
    and I subtracted the free fall acceleration of 607.6 and I ended up with 2802.4
    Is that anything like what I need to do?
     
  9. Nov 9, 2014 #8

    PeroK

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    I don't know where you got that first number. Anyway, first you have to work out what happens when the parachute opens. Hint: Have you converted the initial speed into m/s?
     
  10. Nov 9, 2014 #9

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    I got it by doing mass times gravity, The initial speed is 15.29 m/s
     
  11. Nov 9, 2014 #10

    PeroK

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    That's the right speed, although mass times gravity has nothing to do with it.

    That means the parachutist slows down when the chute opens.

    Can you see how to work out what the deceleration is?
     
  12. Nov 9, 2014 #11

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    Yeah sorry I was answering your question about how I got the first number. Not really, is the deceleration the force of air resistance?
     
  13. Nov 9, 2014 #12

    PeroK

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    You need to calculate the deceleration first.

    Once you have that you calculate the nett force.

    So you need to think carefully about the motion after the chute is opened and about what the 12 m/s represents.
     
  14. Nov 9, 2014 #13

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    Thank you for all your help!
     
  15. Nov 9, 2014 #14

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    So the Deceleration would be 12 m/s and then the net force would be 3.29 m/s?
     
  16. Nov 9, 2014 #15

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    No never mind 3.29 would be the acceleration you use to calculate the net force?
     
  17. Nov 9, 2014 #16

    PeroK

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    You got the right answer but I'm not sure you know exactly why. You can go ahead and calculate the forces now. But you ought to justify that deceleration value.
     
  18. Nov 9, 2014 #17

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    is the end answer 404.86?
     
  19. Nov 9, 2014 #18

    PeroK

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    No. That's less than the force due to gravity. Draw a diagram of the forces and you should see what is happening.
     
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