Question

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?

Answer 1

Part 4

`F_(net) = 8 * 10^4 N`

Step-by-step explanation:

Part 4)

As we know that mass of the Jet is given as

`m = 8 * 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 = (20 - 0)/(20)`

`a = 1 m/s^2`

now net force on the jet is given as

`F_(net) = ma`

`F_(net) = (8 * 10^4)(1)`

`F_(net) = 8 * 10^4 N`

Answer 2

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