An airliner makes an emergency landing with its nose wheel locked in a position perpendicular to its normal rolling position. The forces acting to stop the airliner arise from friction due to the wheels and from the breaking effort of the engines in reverse thrust mode. The force of the engine on the plane is constant, Fengine=−F0. The sum of the horizontal forces on the airliner (in its forward direction) can be written as
F(t)=−F0+((t/ts)−1)F1
from touchdown at time t=0 to the final stop at time ts= 28 s (0≤t≤ts). The mass of the plane is M=80 tonnes (one tonne is 1000 kg). We have F0= 296 kN and F1= 47 kN. Neglect all air drag and friction forces, except the one stated in the problem.
What distance s does the plane go between touchdown and its final stop at time ts? (in meters)
What work do the engines in reverse thrust mode do during the emergency landing?
(magnitude in Joules; the force due to engines is (−F0))
How much heat energy is absorbed by the wheels during the emergency landing?
(magnitude in Joules; the force due to wheels is (F(t)+F0))
To find the distance traveled by the plane between touchdown and its final stop at time t = ts, we need to calculate the area under the graph of the applied force F(t) with respect to time. The integral of F(t) with respect to time will give us the distance traveled.
The integral of F(t) = -F0 + ((t/ts) - 1)F1 with respect to time (t) from 0 to ts will give us the required distance s.
∫[0,ts] [-F0 + ((t/ts) - 1)F1] dt
Integrating the above expression with respect to time from 0 to ts:
= -F0t + (F1(ts^2/2) - F1ts + (F0 - F1)ts - (F1ts^2/2))/ts [evaluating from 0 to ts]
= -F0ts + F1(ts^2)/2 - F1ts + (F0 - F1)ts - (F1ts^2)/2
= -F0ts + (F1ts^2)/2 - F1ts + F0ts - F1ts^2/2
= (F1ts^2)/2 - F1ts^2/2
= 0
Therefore, the distance s traveled by the plane between touchdown and its final stop at time ts is 0 meters.
Now, let's calculate the work done by the engines in reverse thrust mode during the emergency landing. Work is given by the product of force and displacement.
The force exerted by the engines in reverse thrust mode is -F0, and the distance traveled is zero. Hence, the work done by the engines can be calculated as:
Work = Force * Distance
= -F0 * 0
= 0 Joules
Therefore, the work done by the engines in reverse thrust mode during the emergency landing is 0 Joules.
Finally, let's calculate the heat energy absorbed by the wheels during the emergency landing. The force due to the wheels can be calculated as F(t) + F0.
The total work done on the plane can be calculated by integrating the product of the force due to the wheels and the displacement.
The force due to the wheels is (F(t) + F0) = ((t/ts) - 1)F1.
We know that the distance traveled by the plane is zero, so the work done by the wheels can be calculated as:
Work = Force * Distance
= ((t/ts) - 1)F1 * 0
= 0 Joules
Therefore, the heat energy absorbed by the wheels during the emergency landing is 0 Joules.