Respuesta :
Answer:
-41852.8 N
Explanation:
Parameters given:
Final velocity, v = 0km/h = 0m/s
Initial velocity, u = 391.5km/h = 108.75m/s
Time taken to halt, t = 4.5 secs
Weight of the aircraft, W = 16972 N
Force is given as:
F = m*a
Mass, m, can be gotten from weight:
m = W/g = 16972/9.8 = 1731.84 kg
First, we calculate the acceleration:
a = (v - u)/t
a = (0 - 108.75)/4.5 = -108.75/4.5 = -24.2 m/s²
It is a deceleration.
F = 1731.84 * -24.2 = -41852.8N
The force is negative because it needs to act against the motion of the aircraft.
Given Information:
Weight = W = 16,972 N
Initial Velocity = vi = 391.5 kph
Final Velocity = vf = 0 kph
Time = t = 4.5 seconds
Required Information:
Force = F = ?
Answer:
F = -41.85 kN
Explanation:
As we know from Newton's second law of motion
F = ma
Where m is the mass of aircraft and a is acceleration (de-acceleration in this case)
since W = mg
m = W/g
m = 16,972/9.8
m = 1731.83 kg
Now we need to calculate de-acceleration of the aircraft, we know from the kinematics,
a = (vf - vi)/t
first convert km per hour into meter per second
391.5 km/h * 1000 m/3600 s
108.75 m/s
a = (0 - 108.75)/4.5
a = -24.167 m/s²
Finally, we can now calculate the required force
F = ma
F = 1731.83*-24.167
F = -41853.13 N
F = -41.85 kN
Therefore, a force of 41.85 kN is required to bring the aircraft to a safe halt.
