Final answer:
The thrust of each engine on the Boeing 747 as it decelerates from 250 m/s to a stop over a distance of 2500 m is 2.75 x 106 N per engine.
Step-by-step explanation:
The student is asking for the calculation of thrust produced by each engine of a Boeing 747 aircraft as it decelerates and comes to a stop on the runway. To solve this problem, we use the work-energy principle and Newton's second law of motion.
We know the initial kinetic energy (KEi) of the aircraft when it starts to decelerate, the final kinetic energy (KEf) when the aircraft comes to a stop, and the distance (d) over which it stops. The work done by the thrust force (W) is equal to the change in kinetic energy:
W = KEf - KEi
The initial kinetic energy at a speed (v0) of 250 m/s is given by:
KEi = (1/2)mv02
The final kinetic energy is zero since the speed is zero when the aircraft stops.
Thus, W = (1/2)mv02
Substituting the known values:
W = (1/2)(440,000 kg)(250 m/s)2
W = 27.5 x 109 Joules
Now, using the distance (d) over which this work is done, we find the average total thrust (FT), understanding that work is also the product of force and distance:
W = FT * d
FT = W/d
Substituting W and d:
FT = (27.5 x 109 Joules) / (2500 m)
= 11 x 106 N
Since there are four engines, each engine produces a quarter of the total thrust:
Fengine = FT / 4
= (11 x 106 N) / 4
= 2.75 x 106 N per engine