Question

1134567 A jet with mass m = 1.1 Ă— 105 kg jet accelerates down the runway for takeoff at 2 m/s2. 1) What is the net horizontal force on the airplane as it accelerates for takeoff? 2.2*10^5 N Your submissions: 2.2*10^5 Computed value:220000Submitted:Saturday, January 26 at 2:41 PM Feedback:Correct! 2) What is the net vertical force on the airplane as it accelerates for takeoff? 0 N Your submissions: 0 Computed value:0Submitted:Saturday, January 26 at 2:41 PM Feedback:Correct! 3) Once off the ground, the plane climbs upward for 20 seconds. During this time, the vertical speed increases from zero to 21 m/s, while the horizontal speed increases from 80 m/s to 95 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 13 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?

Answer 1

3) The net horizontal force on the airplane as it climbs upward is 8.25 × 10^4 N. 4) The net vertical force on the airplane as it climbs upward is 1.155 × 10^5 N.

Step-by-step explanation:

3) The net horizontal force on the airplane as it climbs upward can be found using the equations of motion. The net force is equal to the mass of the airplane multiplied by its acceleration in the horizontal direction. Since the horizontal speed of the airplane increases from 80 m/s to 95 m/s in 20 seconds, the acceleration can be calculated using the formula:

acceleration = (final velocity - initial velocity) / time

acceleration = (95 m/s - 80 m/s) / 20 s = 0.75 m/s²

Now, the net horizontal force can be calculated using the formula:

net force = mass × acceleration

net force = 1.1 × 10^5 kg × 0.75 m/s² = 8.25 × 10^4 N

Therefore, the net horizontal force on the airplane as it climbs upward is 8.25 × 10^4 N.

4) The net vertical force on the airplane as it climbs upward can be calculated using the same method. The vertical speed of the airplane increases from zero to 21 m/s in 20 seconds, so the vertical acceleration can be calculated as:

acceleration = (final velocity - initial velocity) / time

acceleration = (21 m/s - 0 m/s) / 20 s = 1.05 m/s²

The net vertical force can then be calculated as:

net force = mass × acceleration

net force = 1.1 × 10^5 kg × 1.05 m/s² = 1.155 × 10^5 N

Therefore, the net vertical force on the airplane as it climbs upward is 1.155 × 10^5 N.

Answer 2

Initially jet is moving horizontally

so we know that

`m = 1.1 * 10^5 kg`

`a = 2 m/s^2`

now we have

PART a)

Net horizontal force is given as

`F_x = ma_x`

`F_x = (1.1* 10^5)(2)`

`F_x = 2.2 * 10^5 N`

Part b)

since the acceleration in vertical direction is zero

so as per Newton's law we know

`F = ma`

`F = 0 N`

PART c)

speed of the jet increases in vertical direction from ZERO to 21 m/s in 20 seconds and speed increases in horizontal direction from 95 m/s from 80 m/s in same time

vertical acceleration

`a_y = (21 - 0)/(20) = 1.05 m/s^2`

horizontal acceleration

`a_x = (95 - 80)/(20) = 0.75 m/s^2`

now net horizontal force will be

`F_x = ma_x`

`F_x = (1.1 * 10^5)(0.75) = 8.25 * 10^4 N`

PART d)

Now for vertical force we have

`F_y = ma_y`

`F_y = (1.1 * 10^5)(1.05) = 1.155 * 10^5 N`

PART e)

Now after reaching the cruising level the horizontal speed becomes constant and in vertical direction speed decrease to ZERO from 21 m/s in 13 s

So we have

`a_x = 0`

`a_y = (0 - 21)/(13) = -1.62 m/s^2`

now we have horizontal force as

`F_x = 0`

Part f)

`F_y = ma_y`

`F_y = (1.1 * 10^5)(-1.62) = - 1.78 * 10^5 N`