| 1A-1 : | Kinetic and Potential Energy of an Airplane in Flight | 6 pts |
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| Gravitational acceleration is less at higher altitudes than at sea level. Assume gravitational acceleration as a function of altitude is described by g(m/s2) = 9.806 - 3.2 x 10-6 * h, where h is the altitude (relative to sea level). | |||||||||||||||
| Consider an aircraft flying at 750
km/h at an altitude of 12
km. Before take-off, the aircraft weighed 50 kN at sea
level. Determine its: a.) kinetic energy b.) potential energy relative to sea level. |
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| Read : | Since gravitational acceleration is LESS at higher altitude, the gravitational potential energy of the airplane will not be quite as great as you might ordinarily expect. We need the weight of the airplane at sea level in order to determine the mass of the airplane. We need ot know the mass in order to calculate both the kinetic and gravitation potential energies of the plane. | ||||||||||||||
| Given: | g(m/s2) = a - b * h | v | 750 | km/h | |||||||||||
| a | 9.8066 | m/s2 | h | 12000 | m | ||||||||||
| b | 3.20E-06 | s-2 | Wsea level | 50 | kN | ||||||||||
| gc | 1 | kg-m/N-s2 | |||||||||||||
| Find: | a.) | Ekin | ??? | MJ | b.) | Epot | ??? | MJ | |||||||
| Assumptions: | 1- Gravitational acceleration is a function of altitude only. | ||||||||||||||
| Equations / Data / Solve : | |||||||||||||||
| Let's begin by determining the mass of the airplane from the weight at sea level. | |||||||||||||||
| Newton's 2nd Law of Motion: | 
|
Eqn 1 | |||||||||||||
| or |
|
Eqn 2 | |||||||||||||
| At sea level,
according to the eqn given in the problem statement, the acceleration of gravity is : |
g | 9.8066 | m/s2 | ||||||||||||
| Now, we can solve Eqn 1 for m and plug in values : |
|
Eqn 3 | |||||||||||||
| m | 5099 | kg | |||||||||||||
| Now, we are ready to solve the rest of the problem. | |||||||||||||||
| Part a.) | The definition of kinetic energy is : | 
|
Eqn 4 | ||||||||||||
| Ek | 1.106E+08 | J | |||||||||||||
| Ek | 111 | MJ | |||||||||||||
| Part b.) | The definition of gravitational potential energy, relative to sea level, is : | 
|
Eqn 5 | ||||||||||||
| The problem is what value of g do we use ? Do we simply use g at the altitude of the plane? Or do we use some sort of average value of g ? | |||||||||||||||
| Let's think about this part a bit more carefully. | 
|
Eqn 6 | |||||||||||||
| The differential
increase in the potential energy of an object infinitessimally above sea level is: |
|
Eqn 7 | |||||||||||||
| So, the gravitational
potential energy of an oject that is a distance h above sea level is : |
|
Eqn 8 | |||||||||||||
| Now, if m, g and gc are all constants, Eqn 7 simply reduces to Eqn 5 because the integral of dx from 0 to h is just h. In our problem, however, g is NOT a constant. Therefore : | |||||||||||||||
|
Eqn 9 | ||||||||||||||
| Epot | 5.988E+08 | J | Epot | 598.8 | MJ | ||||||||||
| So, what "average" value of g should we have used in Eqn 5 ? Let's combine Eqns 5 and 8 and see what we get. | |||||||||||||||
|
or : | 
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Eqn 10 | ||||||||||||
| So, the average effective value of the gravitational acceleration for determining the potential energy of the airplane in this problem is equal to the gravitational acceleration at HALF of the actual altitude of the airplane. Would this be true if g = a - b x -c x2 ?? Nope. What is special about the equation for g given in this problem that leads to the interesting result in Eqn 10 ? | |||||||||||||||
| Answers : | a.) | Ek | 111 | MJ | b.) | Epot | 599 | MJ | |||||||
