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 horizontal component of the net force on the airplane as it accelerates for takeoff?

2) What is the vertical component of the net 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 29 m/s, while the horizontal speed increases from 80 m/s to 96 m/s.

What is the horizontal componenet of the net force on the airplane as it climbs upward?

4) What is the vertical componenet of the net 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 horizontal component of the net force on the airplane during leveling off?

6) What is the vertical component of the net force on the airplane during leveling off?

Explanation / Answer

1) Horizontal component is:

Fx = max = (5 × 10)(1.8) =9× 10 N

2) takeoff the motion is in the horizontal so there is no acceleration in the vertical axis so:

Fy = 0

3) The horizontal component is:

Fx = max

where:

ax = (vxf-vxi)/t = (96-80)/(20)=0.8 m/s²

so: Fx = 5x10(0.8)=4x10 m/s²

4) Fy = may

where:

ay =(Vyf-Vyi)/t = (29-0)/20 = 1.45 m/s²

Fy = 5x10 (1.45) = 7.25 x10 N

5) As the horizontal speed is constant there is no net force

Fx = 0 N

6) In this case

ay = (29-0)/12 = 2.42 m/s²

Fy = may = 5x 10 (2.42) = 1.21x10 N

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