A helicopter of mass \(2000\) kg rises with a vertical acceleration of \(15~\mathrm{m/s^2}\). The total mass of the crew and passengers is \(500\) kg. The magnitude of the force on the floor of the helicopter by the crew and passengers is: (Take \(g=10~\mathrm{m/s^2}\))
1. \(12500~\mathrm{N}\)
2. \(1250~\mathrm{N}\)
3. \(50000~\mathrm{N}\)
4. \(5000~\mathrm{N}\)

Hint: Recall pseudo force.
Step 1: Apply pseudo force and find force on the floor of the helicopter by the crew and passengers.
Given, mass of helicopter m1 = 2000 kg
Mass of the crew and passengers m2 = 500 kg
Acceleration in the vertical direction a=15m/s2() and g=10m/s2()
(a) Force on the floor of the helicopter by the crew and passengers
m2(g+a)=500(10+15)N500×25N=12500N
Step 2: Find action of the helicopter on the surrounding air.
(b) Action of the rotor of the helicopter on the surrounding air =(m1+m2)(g+a)
               = (2000 + 500) X (10 + 15) = 2500 x 25
               = 62500N (downward)
Step 3: Find force on the helicopter due to the surrounding air.
(c) Force on the helicopter due to the surrounding air
               = reaction of force applied by helicopter
               = 62500 N (upward).