A jet with mass m = 8 × 104 kg jet accelerates down the runway for takeoff at 1.5 m/s2.

I have Found out so far that...

1) net horizontal force(takeoff) = 120000

2)Net vertical force(takeoff) = 0

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

3)What is the net horizontal force on the airplane as it climbs upward = 64000

I need help here...

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

The net vertical force for this jet that is accelerating from 0 m/s to 20 m/s over 20 seconds is:

F=ma=m(V2-V1)/(t2-t1)

where V2 is the final speed and V1 is the initial speed and t2 is the final time and t1 is the initial time.

F=(8x10^4kg)(20m/s - 0 m/s)/20s =(8x10^4kg)(1m/s^2) = 8x10^4 N

Answer Link

aristocles aristocles · Answer 2 · 2019-08-21T20:05:00-04:00

Answer:

Part 4

\(F_{net} = 8 \times 10^4 N\)

Explanation:

Part 4)

As we know that mass of the Jet is given as

\(m = 8 \times 10^4 kg\)

now in vertical direction the speed of the jet is changing from zero to 20 m/s

so we can say

\(v_i = 0\)

\(v_f = 20 m/s\)

now time taken to change the speed is given as

\(\Delta t = 20 s\)

now we can find the acceleration of the jet

\(a = \frac{20 - 0}{20}\)

\(a = 1 m/s^2\)

now net force on the jet is given as

\(F_{net} = ma\)

\(F_{net} = (8 \times 10^4)(1)\)

\(F_{net} = 8 \times 10^4 N\)