A jet with mass m = 10^5 kg jet accelerates down the runway for takeoff at 1.8 m

Question

A jet with mass m = 10^5 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

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

3) Once off the ground, the plane climbs upward for 20 seconds. During this time, the vertical speed increases from zero to 16 m/s, while the horizontal speed increases from 80 m/s to 92 m/s. What is the net horizontal force on the airplane as it climbs upward?____________ N

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

5)After reaching cruising altitude, the plane levels off, keeping the horizontal speed constant, but smoothly reducing the vertical speed to zero, in 17 seconds.

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

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

Explanation / Answer

A jet with mass m = 10^5 kg jet accelerates down the runway for takeoff at 1.8 m/s2.

1) net horizontal force = 10^5 x 1.8 =1.8x10^5 N.

2) net vertical force is zero. (because there is no vertical acceleration)

Once off the ground, the plane climbs upward for 20 seconds. During this time, the vertical speed increases from zero to 16 m/s, while the horizontal speed increases from 80 m/s to 92 m/s.

3)net horizontal force = 10^5x(16-0)/20 = 8x10^4 N

4)net vertical force = 10^5x(92-80)/20 = 6x10^4 N.

5) After reaching cruising altitude, the plane levels off, keeping the horizontal speed constant, but smoothly reducing the vertical speed to zero, in 17 seconds.

net horizontal force= 0 .(speed is constant)

6)net vertical force = 10^5 x(0-16)/17 = 9.4 x 10^4 N

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