As given that, mass of helicopter $\left(m_1\right)=2000 \mathrm{~kg}$
Mass of the crew and passengers $m_2=500 \mathrm{~kg}$
Acceleration (of helicopter) in vertical direction
$a=15 \mathrm{~m} / \mathrm{s}^2(\uparrow)$ and $g=10 \mathrm{~m} / \mathrm{s}^2(\downarrow)$
(a) Force $\left(F_1\right)$ on the floor of the helicopter by the crew and passengers will be equal to apparent -weight (left).
$$
\begin{gathered}
F_1=m_2(g+a)=500(10+15) \mathrm{N} \\
=500 \times 25 \mathrm{~N}=12500 \mathrm{~N}
\end{gathered}
$$
(b) Action of the rotor of the helicopter on the surrounding air will be equal to the reaction force $\left(F_2\right)$ by Newton's third law due to which helicopter along with crew and passenger rises up to acceleration
$$
\begin{aligned}
&F_2=\left(m_1+m_2\right)(g+a) \\
&=(2000+500) \times(10+15)=2500 \times 25 \\
&=62500 \mathrm{~N} \text { (downward) }
\end{aligned}
$$
(c) Force $\left(F_3\right)$ acting on the helicopter (by reaction force) due to the surrounding air
= reaction of force applied by helicopter
$$
F_3=62500 \mathrm{~N} \text { (upward), }
$$
as Action force
$$
F_2=-62500 \mathrm{~N} \text { (downward). }
$$