A jet with mass m = 5 × 10 4 kg jet accelerates down the runway for takeoff at 1

Question

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

1)

What is the net horizontal force on the airplane as it accelerates for takeoff?

N

Your submissions:

2)

What is the net vertical force on the airplane as it accelerates for takeoff?

N

Your submissions:

3)

Once off the ground, the plane climbs upward for 20 seconds. During this time, the vertical speed increases from zero to 15 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?

N

Your submissions:

4)

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

N

Your submissions:

5)

After reaching cruising altitude, the plane levels off, keeping the horizontal speed constant, but smoothly reducing the vertical speed to zero, in 13 seconds.

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

N

Your submissions:

6)

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

N

Your submissions:

(Survey Question)

7)

Below is some space to write notes on this problem

Explanation / Answer

An 50000 kg jet accelerates down the runway for takeoff at 1.8 m/s2.
Force = mass * acceleration
Force = 5.0 * 10^4 * 1.8 = 90000 N

I found the net horizontal force on the airplane as it accelerates for takeoff to be 90000 N and the net vertical force of that to be 0 N.
The next part of the question states: Once off the ground, the plane climbs upward for 20 seconds. During this time, the vertical speed increases from zero to 15 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 = 5.0 * 10^4 * 0.8 = 6.3 * 10^4 N

Net force = 1.12 * 10^5 + 6.3 * 10^4 =40000 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 15 m/s
Vertical acceleration = velocity ÷ time = +15 ÷ 20 = +0.75 m/s^2
Force causing acceleration = 5.0 * 10^4 * 0.75 = +37500

To accelerate the airplane at +0.75 m/s^2 vertically, the net upward force must = weight of plane + Force causing acceleration.

Force = 5.0 * 10^4 * 9.8 + +37500
Force = +527500 N
This is the net vertical force on the airplane as it to accelerate upward at +0.75 m/s^2.

Next, the vertical speed is reduced from 15 m/s to 0 m/s, in 13 seconds.
Acceleration = velocity ÷ time = -15 ÷ 13 = -(15/13) m/s^2
The vertical force that caused the -(25/13) m/s^2 acceleration =
5.0 * 10^4 * -(15/13) = -57692.30 N

Net vertical force on the airplane as it levels off =
+527500 + -57692.30 = 469807.7 N

The weight = 5.0 * 10^4 * 9.8 = 490000 N
To have a vertical velocity of 0 m/s the upward vertical force on the plane must equal its weight!!!


So the plane will not level off at 0 m/s upward, unless the plane increases the vertical force from469807.7 N to 490000 N

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