A jet with mass m = 5 × 104 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?
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 29 m/s, while the horizontal speed increases from 80 m/s to 96 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 12 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?
To find the net horizontal force and net vertical force on the airplane in different scenarios, we need to apply Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
1) Net Horizontal Force during takeoff:
Given:
Mass of the airplane (m) = 5 × 10^4 kg
Acceleration (a) = 1.8 m/s^2
The net horizontal force (F) can be calculated using the formula:
F = m * a
Substituting the given values, we have:
F = 5 × 10^4 kg * 1.8 m/s^2
F = 9 × 10^4 N
Therefore, the net horizontal force on the airplane during takeoff is 9 × 10^4 Newtons.
2) Net Vertical Force during takeoff:
Since the question only asks for the net horizontal and vertical forces, we are not given any specific information about the net vertical force during takeoff. However, we can say that the net vertical force would be the total weight of the airplane acting downward, which can be calculated as:
Weight (W) = m * g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
So, the net vertical force during takeoff is simply the weight of the airplane.
3) Net Horizontal Force as the airplane climbs upward:
Given:
Initial horizontal speed (v1) = 80 m/s
Final horizontal speed (v2) = 96 m/s
Time (t) = 20 s
To calculate the net horizontal force during this time, we need the change in horizontal momentum (Δp). The change in momentum is given by the formula:
Δp = m * (v2 - v1)
Substituting the given values, we have:
Δp = 5 × 10^4 kg * (96 m/s - 80 m/s)
Δp = 5 × 10^4 kg * 16 m/s
Δp = 8 × 10^5 kg·m/s
The net horizontal force (F) can then be calculated using the formula:
F = Δp / t
Substituting the given values, we have:
F = (8 × 10^5 kg·m/s) / 20 s
F = 4 × 10^4 N
Therefore, the net horizontal force on the airplane as it climbs upward is 4 × 10^4 Newtons.
4) Net Vertical Force as the airplane climbs upward:
During the upward climb, we know that the vertical speed increases from 0 m/s to 29 m/s. Since we are not given any information about the time required for this change, we cannot determine the net vertical force specifically for this context.
5) Net Horizontal Force as the airplane levels off:
Since the horizontal speed is kept constant as the airplane levels off, there is no change in horizontal momentum. Therefore, the net horizontal force is zero.
6) Net Vertical Force as the airplane levels off:
The vertical speed is reduced from 29 m/s to 0 m/s over a period of 12 seconds. To calculate the net vertical force during this time, we need the change in vertical momentum.
Δp = m * (v2 - v1)
Δp = 5 × 10^4 kg * (0 m/s - 29 m/s)
Δp = -1.45 × 10^6 kg·m/s (since the velocity change is negative)
The net vertical force (F) can then be calculated using the formula:
F = Δp / t
F = (-1.45 × 10^6 kg·m/s) / 12 s
F = -1.21 × 10^5 N
Therefore, the net vertical force on the airplane as it levels off is -1.21 × 10^5 Newtons, directed upward (since the force needed to decrease the vertical speed is opposite to the weight acting downward).