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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?

Respuesta :

taskmasters
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:

Part 4

[tex]F_{net} = 8 \times 10^4 N[/tex]

Explanation:

Part 4)

As we know that mass of the Jet is given as

[tex]m = 8 \times 10^4 kg[/tex]

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

so we can say

[tex]v_i = 0[/tex]

[tex]v_f = 20 m/s[/tex]

now time taken to change the speed is given as

[tex]\Delta t = 20 s[/tex]

now we can find the acceleration of the jet

[tex]a = \frac{20 - 0}{20}[/tex]

[tex]a = 1 m/s^2[/tex]

now net force on the jet is given as

[tex]F_{net} = ma[/tex]

[tex]F_{net} = (8 \times 10^4)(1)[/tex]

[tex]F_{net} = 8 \times 10^4 N[/tex]

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