$A$ helicopter of mass $1000 \;kg$ rises with a vertical acceleration of $15\; m s^{-2}$. The crew and the passengers weigh $300\; kg$. Give the magnitude and direction of the
$(a)$ force on the floor by the crew and passengers,
$(b)$ action of the rotor of the helicopter on the surrounding air,
$(c)$ force on the helicopter due to the surrounding air.

(A-D) Mass of the helicopter,$m_h = 1000 \; kg$. Mass of the crew and passengers,$m_p = 300 \; kg$.
Total mass of the system,$m = 1300 \; kg$.
Acceleration of the helicopter,$a = 15 \; m s^{-2}$.
$(a)$ Using Newton's second law of motion for the crew and passengers,the normal force $R$ exerted by the floor is:
$R - m_p g = m_p a \implies R = m_p(g + a)$
$R = 300(10 + 15) = 300 \times 25 = 7500 \; N$.
By Newton's third law,the force on the floor by the crew and passengers is $7500 \; N$,directed downward.
$(b)$ Using Newton's second law for the entire system (helicopter + crew + passengers):
$F_{air} - mg = ma \implies F_{air} = m(g + a)$
$F_{air} = 1300(10 + 15) = 1300 \times 25 = 32500 \; N$.
By Newton's third law,the action of the rotor on the surrounding air is $32500 \; N$,directed downward.
$(c)$ The force on the helicopter due to the surrounding air is $32500 \; N$,directed upward.