Question

A jet with mass m = 8 × 10⁴ kg jet accelerates down the runway for takeoff at 1.5 m/s^2. Once off the ground, the plane climbs upward for 20 seconds. During this time, the vertical speed increases from zero to 21 m/s, while the horizontal speed increases from 80 m/s to 96 m/s. After reaching cruising altitude, the plane levels off, keeping the horizontal speed constant, but smoothly reducing the vertical speed to zero, in 11 seconds.What is the net vertical force on the airplane as it levels off?

Answer 1

An 80000 kg jet accelerates down the runway for takeoff at 1.5 m/s2.
Force = mass * acceleration
Force = 8.0 * 10^4 * 1.5 = 1.2 * 10^5 N

Once off the ground, the plane climbs upward for 20 seconds. During this time, the vertical speed increases from zero to 21 m/s, while the horizontal speed increases from 80 m/s to 96 m/s.

Horizontal acceleration = ∆velocity ÷ time = 16 ÷ 20 = 0.8 m/s^2
Increase in horizontal force = mass * acceleration
Increase in horizontal force = 8.0 * 10^4 * 0.8 = 6.4 * 10^4 N

Net force = 1.2 * 10^5 + 6.4 * 10^4 = 1.84 * 10^5 N
This is the net horizontal force on the airplane as it climbs upward for 20 seconds. This net horizontal force remains constant!

During the 20 seconds, the vertical speed increases from zero to 21 m/s
Vertical acceleration = ∆velocity ÷ time = +21 ÷ 20 = +1.05 m/s^2
Force causing acceleration = 8.0 * 10^4 * 1.05 = +8.4 * 10^4

To accelerate the airplane at +1.05 m/s^2 vertically, the net upward force must = weight of plane + Force causing acceleration.

Force = 8.0 * 10^4 * 9.8 + +8.4 * 10^4
Force = +8.68 * 10^5 N
This is the net vertical force on the airplane as it to accelerate upward at +1.05 m/s^2.

Next, the vertical speed is reduced from 21 m/s to 0 m/s, in 11 seconds.
Acceleration = velocity ÷ time = -21 ÷ 11 = -(21/11) m/s^2
The vertical force that caused the -(21/11) m/s^2 acceleration =
8.0 * 10^4 * -(21/11) = -1.52727 * 10^5 N

Net vertical force on the airplane as it levels off =
+8.68 * 10^5 + -1.52727 * 10^5 = 7.1527 * 10^5 N ………..Ans.