A damaged S-3A Viking is trying to land on an aircraft carrier. It cannot deploy
ID: 1816651 • Letter: A
Question
A damaged S-3A Viking is trying to land on an aircraft carrier. It cannot deploy its tailhook and must use the emergency barricade system instead of the arrestor cables normally used for landing aircraft on the carrier. The aircraft has a mass of 17000 kg and approaches the landing area at 240 km per hour. If the barricade can be compared to a rubber band with a spring constant, k, of 1.8 kN/m:a. How far will the plane travel, assuming the pilot has no brakes or rear thrust, the barricade can stretch without breaking, and it completely stops the plane?
b. If the barricade has an efficiency of 75%, and the landing area stretches 280 m beyond the initial position of the barricade, will the plane make it? Assume the barricade can stretch without breaking.
c. If the plane's brakes are functional, and the efficiency of the barricade and the brakes is 80%, how much energy will the brakes have to supply to assist the barricade and stop the plane in 150 m?
Explanation / Answer
Sorry that the equations are not done in the editor, but for some reason it kept crashing, and I wanted to answer your question as soon as possible.
Conservation of energy is a useful approach to this problem. Consider the plane when it first touches down to be point 1, and the point at which the plane has come to rest to be point 2. Assuming no frictional losses, this system can then be described by the equation
PE1+KE1=PE2+KE2 (1)
where PE is the potential energy, KE is the kinetic energy, and the subscripts correspond to the state of the system being observed. Because the vertical height of the plane will not change between these two points, the potential energy of plane can be considered to be zero throughout the process. Additionally, assuming that the barricade is not stretched or compressed prior to coming in contact with the plane, its initial potential energy can be considered to be zero as well. As the final position is static, the sum of the final kinetic energies will be identically equal to zero.
Therefore, the only energies that need to be considered for this problem, are the initial kinetic energy of the plane, and the final potential energy stored in the spring. With the assumption of an ideal system, these two values can be set equal to each other such that the equation
KE1 = PE2 (2)
is true. Knowing that
KE1 = 1/2 mv2 (3)
where m is the mass of the plane, v is the velocity of the plane, and
PE2 = 1/2 Kx2 (4)
where K is the assumed spring constant, and x is the displacement of the barricade. Making these substitutions, the modeling equation becomes.
1/2 mv2 = 1/2 Kx2 (5)
Using the constants m = 17000 kg, v = 66.67 m/s (1 km = 103 m, 1 hr = 3600 s), K = 1800 N/m (1 kN = 103 m), x is found to be equal to 204.9 m.
a. 204.9 m (ans)
For part b, the barricade is asssumed to be 75 % efficient. The easiest way to consider this, is to note that the efficiency of a spring is equal to the energy in divided by the energy out. In this case,
= KE1 / PE2 (6)
where is the efficiency. Making the same substitutions as before the equation becomes:
= mv2 / Kx2 (7)
Using the constants m = 17000 kg, v = 66.67 m/s (1 km = 103 m, 1 hr = 3600 s), K = 1800 N/m (1 kN = 103 m), and = .75, x is found to be 236.6 m. Knowing that the barricade can stretch 280 m without stretching, this shows that the plane will come to a stop.
b. yes, the plane will come to a stop after traveling 236.6 m (ans)
For part c, the barricade and brakes around both assumed to be 80 % efficient. This is similar to part b, except there is an added input energy, q, that is taking away from the kinetic energy of state 1. Therefore
= (mv2 - q ) / Kx2 (8)
Using the constants m = 17000 kg, v = 66.67 m/s (1 km = 103 m, 1 hr = 3600 s), K = 1800 N/m (1 kN = 103 m), = .8, and x = 150 m, the energy released from the system due to the brakes is found to be 4.32*107 J. However, this is the value of the energy that affects the system, but not necessarily the energy utilized by the brakes. Taking into account the 80 % efficiency of the brakes yields
4.32*107 J / .8 = 5.39*107 J (9)
c. 5.39*107 J is the energy used by the brakes to stop the system in 150 m.