Acceleration of the helicopter as it starts to rise

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SUMMARY

The discussion focuses on calculating the acceleration of a helicopter as it begins to rise, given that it propels 2400 kg of air downwards at a speed of 11 m/s. The net force acting on the helicopter is determined to be 26,400 N, leading to the equation 26,400 - 2500(9.81) = (2500)a. The correct acceleration is calculated to be 9.58 m/s², despite an initial miscalculation that suggested 0.75 m/s². The confusion arose from a potential error in mass values used in the calculations.

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anlenemilk
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Homework Statement


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



Homework Equations



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

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

What textbook is this problem from?
 
anlenemilk said:
26400-2500(9.81)=(2500)a
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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