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Q. A helicopter of mass $1000\, kg$ rises with a vertical acceleration of $15 \,ms ^{-2} .$ The crew and the passengers weigh $300\, kg$. The magnitude of the action of the rotor of the helicopter on the surrounding air is

Laws of Motion

Solution:

The required action is equal and opposite to the force on the helicopter due to the surrounding air.
$F-m g=m a$
$F=m(g+a)$
$=(1000 ] \,kg +300\, kg )\left(10 \, ms ^{-2}+15 \, ms ^{-2}\right)$
$=(1300)(25) N =3.25 \times 10^{4} N$