A jet with mass m = 5 × 104 kg jet accelerates down the runway for takeoff at 1.8 m/s2.
and ...
1)What is the net horizontal force on the airplane as it accelerates for takeoff?
2)What is the net vertical force on the airplane as it accelerates for takeoff?
3)Once off the ground, the plane climbs upward for 20 seconds. During this time, the vertical speed increases from zero to 28 m/s, while the horizontal speed increases from 80 m/s to 95 m/s.
4)What is the net vertical force on the airplane as it climbs upward?
5)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?
6)What is the net vertical force on the airplane as it levels off?
To understand the motion of a jet during takeoff, we can apply Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. The formula can be written as:
Force (F) = mass (m) × acceleration (a)
In this case, we are given the mass of the jet (m = 5 × 10^4 kg) and the acceleration (a = 1.8 m/s^2). We want to find the force acting on the jet during takeoff.
Using the given values, we can calculate the force using the formula:
F = m × a
F = (5 × 10^4 kg) × (1.8 m/s^2)
F = 9 × 10^4 kg·m/s^2
The force acting on the jet during takeoff is 9 × 10^4 kg·m/s^2.