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Acceleration of the helicopter as it starts to rise!

  1. Sep 2, 2012 #1
    1. The problem statement, all variables and given/known data
    At take off, the rotor blades of a helicopter propel 2400 kg of air vertically downwards each second. The air, intially at rest, is given a speed of 11ms^-1
    QN: the mass of the helicopter is 2500 kg. Find the acceleration of the helicopter as it starts to rise



    2. Relevant equations

    <F>=Change in momentum /change in time
    Fnet=ma

    3. The attempt at a solution

    <F>=Change in momentum /change in time
    = (2400*11-0)/1= 26400N
    26400-2500(9.81)=(2500)a
    a=0.75ms^-2

    The answer is 9.58ms^-2
    I have no idea how to get this answer! HELP ME! THANK YOU VERY MUCH
     
  2. jcsd
  3. Sep 2, 2012 #2

    Doc Al

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    Staff: Mentor

    I'd say that your answer is correct and the given answer is wrong.

    What textbook is this problem from?
     
  4. Sep 2, 2012 #3

    Doc Al

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    Staff: Mentor

    Note that you'll get the given answer if you changed one of the masses to 250 (which is clearly wrong):
    26400-250(9.81)=(2500)a​
    That could be how the wrong answer was attained.
     
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