Question

A jet with mass m=120000kg jet accelerates down the runway for takeoff at 2.2m/s^2. Once off the ground, the plane climbs upward for 20seconds. During this time, the vertical speed increases from zero to 27 m/s, while the horizontal speed increases from 80m/s to 90 m/s.After reaching cruising altitude, the plane levels off, keeping the horizontal speed constant, but smoothly reducing the vertical speed to zero, in 17 seconds. 1) What is the net horizontal force on the airplane as it levels off?2) What is the net vertical force on the airplane as it levels off?

Answer 1

Given data:

Mass of the jet;

`m=120000\text{ kg}`

Initial horizontal velocity of the jet while levels off,

`u_x=90\text{ m/s}`

Final horizontal velocity of the jet while levels off,

`v_y=90\text{ m/s}`

While the levels off the horizontal velocity is constant.

Initial vertical velocity of the jet while levels off,

`u_y=27\text{ m/s}`

Final vertical velocity of the jet while levels off,

`v_y=0`

Time;

`t=17\text{ s}`

Part (1),

The horizontal acceleration of the jet while levels off is given as,

`a_x=(v_x-u_x)/(t)`

Substituting all known values,

`\begin{gathered} a_x=\frac{(90\text{ m/s})-(90\text{ m/s})}{17\text{ s}} \\ =0 \end{gathered}`

The net horizontal force on the airplane as it levels off is given as,

`F_x=ma_x`

Substituting all known values,

`\begin{gathered} F_x=(120000\text{ kg})*0 \\ =0\text{ N} \end{gathered}`

Therefore, the net horizontal force on the airplane as it levels off is 0 N.

Part (2)

The vertical acceleration of the jet while levels off is given as,

`a_y=(v_y-u_y)/(t)`

Substituting all known values,

`\begin{gathered} a_y=\frac{(0)-(27\text{ m/s})}{17\text{ s}} \\ \approx-1.59\text{ m/s}^2 \end{gathered}`

The net vertical force on the airplane as it levels off is given as,

`F_y=ma_y`

Substituting all known values,

`\begin{gathered} F_y=(120000\text{ kg})*(-1.59\text{ m/s}^2) \\ =190800\text{ N} \end{gathered}`

Therefore, the net vertical force on the airplane as it levels off is 190800 N.