In this problem, we're going to use Newton's second law:

$\overline{){\mathbf{\Sigma}}{\mathbf{F}}{\mathbf{=}}{\mathbf{m}}{\mathbf{a}}}$

We'll also use the kinematic equations:

$\overline{)\mathbf{}{{\mathit{v}}}_{{\mathit{f}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{{\mathit{v}}}_{{\mathbf{0}}}{\mathbf{}}{\mathbf{+}}{\mathit{a}}{\mathit{t}}}$

**1)**

From Newton's second law of motion:

$\begin{array}{rcl}\mathbf{F}& \mathbf{=}& \mathbf{m}\mathbf{a}\\ & \mathbf{=}& \mathbf{(}\mathbf{100000}\mathbf{.}\mathbf{0}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{9}\mathbf{)}\end{array}$

F = 190000 N

The net horizontal force on the airplane as it accelerates for takeoff is 190000 N.

A jet with mass m = 100000.0 kg jet accelerates down the runway for takeoff at 1.9 m/s^{2}.

1) What is the net horizontal force on the airplane as it accelerates for takeoff?

2) What is the net vertical 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 17.0 m/s, while the horizontal speed increases from 80.0 m/s to 93.0 m/s. What is the net horizontal force on the airplane as it climbs upward?

4) What is the net vertical 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 11.0 seconds. What is the net horizontal force on the airplane as it levels off?

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

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