A jet with mass m = 1.3 × 105 kg jet accelerates down the runway for takeoff at

Question

A jet with mass m = 1.3 × 105 kg jet accelerates down the runway for takeoff at 1.5 m/s2. 1) What is the net horizontal force on the airplane as it accelerates for takeoff? 1.274*10^6 N Submit 2) What is the net vertical force on the airplane as it accelerates for takeoff? 1.469*10^6 N Submit 3) Once off the ground, the plane climbs upward for 20 seconds. During this time, the vertical speed increases from zero to 19 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 Submit 4) What is the net vertical force on the airplane as it climbs upward? N Submit 5) After reaching cruising altitude, the plane levels off, keeping the horizontal speed constant, but smoothly reducing the vertical speed to zero, in 13 seconds. What is the net horizontal force on the airplane as it levels off? N Submit 6) What is the net vertical force on the airplane as it levels off?

Explanation / Answer

Given that

A jet with mass m = 1.3 × 105 kg

Jet accelerates down the runway for takeoff at(a) = 1.5 m/s2.

1)

Now the net horizontal forcce is (F) =ma =1.3*105*1.5 =195000N

2)

The net vertical force on the plane is zero,because there is no acceleration in the vertical direction i.e weight is balanced by the normal reaction on the tyres.

3)

Now the net horizontal force ont he airplane as it climbs up is F =mv/t =(1.3*105)(12)/20 =78000N

4)

Now the net vertical force is F =m(v2-v1)/t =(1.3*105)(19m/s)/20 =123500N

5)

Since horizontal speed is not changing,Net force in horizontal direction will be 0

6)

For the vertical direction vf =vi+at

                                  0 =19m/s+a(13m/s)

a =-1.461m/s2

Now net vertical force on the airplane as it levels off is

F =ma =(1.3 × 105 kg )(-1.461m/s2) =190000N

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