Question

A jet with mass m = 5 × 10⁴ kg jet accelerates down the runway for takeoff at 1.8 m/s².

a) Once off the ground, the plane climbs upward for 20 seconds. During this time, the vertical speed increases from zero to 29 m/s, while the horizontal speed increases from 80 m/s to 96 m/s.

What is the net horizontal force on the airplane as it climbs upward?

b) What is the net vertical force on the airplane as it climbs upward?

c) After reaching cruising altitude, the plane levels off, keeping the horizontal speed constant, but smoothly reducing the vertical speed to zero, in 12 seconds.

What is the net horizontal force on the airplane as it levels off?

d)  What is the net vertical force on the airplane as it levels off?

Answer 1

a)

The acceleration acting horizontally will be

Then horizontal force acting on it in this time will be

F_H=ma =(5*10⁴)(0.8)=4*10⁴ N

b)

The net acceleration acting vertically on plane,

The vertical force acting on it in this time will be

F_V=ma_v=(5*10⁴)(1.45)=7.25*10⁴ N

c)

Since horizontal speed is constant ,Therefore

a_H=0

the net horizontal force on the airplane as it levels off

F_H=ma_H=0 N

d)

Then vertical force acting on it in this time will be

F_V=ma_V =(5*10⁴)(-2.4167) =-1.2083*10⁵ N