Question

An aircraft weighing 16,972 N returns to the aircraft carrier and lands. It goes from 391.5 kilometers per hour (kph) to 0 kph in 4.5 seconds. What is the force required to bring the aircraft to a safe halt

Answer 1

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.

Answer 2

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.