Steven W. answered • 11/18/16
Tutor
4.9
(4,337)
Physics Ph.D., college instructor (calc- and algebra-based)
Hi Dominic!
(a) Presuming that the plane starts on the ground (h=0) at rest, then its initial energy is 0 (no kinetic energy; no gravitational potential energy)
Therefore, it must gain enough kinetic energy to end up at 300 m/s, and enough gravitational potential energy to end up 12,000 meters higher than where it started. This means the total (mechanical) energy needed is:
ME = KE + PE = (1/2)mv2 + mgh
where
v = final speed = 300 m/s
h = 12,000 m
With those numbers, and the 1.5 x 105-kg mass of the plane, we have:
ME = (1/2)(1.5x105 kg)(300 m/s)2 + (1.5x105 kg)(9.8 m/s2)(12,000 m) = 2.44 x 1010 J
If the engine produces 100 MW (= 1.0x108 W = 1.0x108 J/s). then, to produce 2.44 x 1010 J (presuming all the power goes to propulsion), it would take:
2.44 x 1010 J * (1 s)/(1.0 x 108 J) = 244 s
(b) If it actually takes 950 s, then the power will be the total energy produced (which will be the same as before, 2.44 x 1010 J) divided by this longer time, since power is energy delivered (or work done) per unit time.
(a) Presuming that the plane starts on the ground (h=0) at rest, then its initial energy is 0 (no kinetic energy; no gravitational potential energy)
Therefore, it must gain enough kinetic energy to end up at 300 m/s, and enough gravitational potential energy to end up 12,000 meters higher than where it started. This means the total (mechanical) energy needed is:
ME = KE + PE = (1/2)mv2 + mgh
where
v = final speed = 300 m/s
h = 12,000 m
With those numbers, and the 1.5 x 105-kg mass of the plane, we have:
ME = (1/2)(1.5x105 kg)(300 m/s)2 + (1.5x105 kg)(9.8 m/s2)(12,000 m) = 2.44 x 1010 J
If the engine produces 100 MW (= 1.0x108 W = 1.0x108 J/s). then, to produce 2.44 x 1010 J (presuming all the power goes to propulsion), it would take:
2.44 x 1010 J * (1 s)/(1.0 x 108 J) = 244 s
(b) If it actually takes 950 s, then the power will be the total energy produced (which will be the same as before, 2.44 x 1010 J) divided by this longer time, since power is energy delivered (or work done) per unit time.
Presumably, then, this is the effective power provided by the engine (without assuming, as in (a), that all its power goes to propulsion)
Peff = (2.44x1010 J)/(950 s) = 25.6 MW
Also, assuming the force provided by the engines is in the direction of the plane's displacement, we can now compute the force provided by just the engines, knowing that the work of this force must provide the full change in energy over this motion (assuming no air resistance).
We can determine the displacement of the plane, assuming constant acceleration, using kinematics. In this case, we want to determine displacement, and we know its initial velocity, final velocity, and the time of the motion. So we may use the equation:
d = (1/2)(vo+v)t
where
vo = initial velocity, = 0 m/s
v = final velocity = 300 m/s
t = 1200 s
so
d = (1/2)(0+300 m/s)(1200 s) = 1.8 x 105 m
Also, assuming the force provided by the engines is in the direction of the plane's displacement, we can now compute the force provided by just the engines, knowing that the work of this force must provide the full change in energy over this motion (assuming no air resistance).
We can determine the displacement of the plane, assuming constant acceleration, using kinematics. In this case, we want to determine displacement, and we know its initial velocity, final velocity, and the time of the motion. So we may use the equation:
d = (1/2)(vo+v)t
where
vo = initial velocity, = 0 m/s
v = final velocity = 300 m/s
t = 1200 s
so
d = (1/2)(0+300 m/s)(1200 s) = 1.8 x 105 m
Presumably, Part (b) gave us information about how much work the engines alone would do to get the plane to that final state, in the absence of air resistance (note: the work done my gravity is included by putting in the change in gravitational potential energy as part of the overall change in mechanical energy). We can use that to compute the force the engines exert on the plane, which will be the same even when air resistance is added.
To calculate the force of the engines on the plane alone, we can redo the previous calculation, except using 950 s at t instead of 1200 s (all other quantities remain the same, since the plane still has to reach the same final energy state). This gives:
d = 1.43 x 105 m
Then, the work done by the engines, in terms of the force-displacement definition (assuming the engine force is always in the direction of motion) is:
We = Fed = Fe(1.42 x 105 m) = 2.44 x 1010 J (since, as before, the work done by the engines, in this case, equals the (mechanical) energy generated to put the plane in the indicated final state)
Thus, we can solve for the force of the engines, which is Fe = 1.71 x 105 N = 0.17 kN
Since, even with the air resistance, the plane still ends up in the same final state, we can say that the net work done with air resistance is the same as the net work done without it. But this time, there are two forces doing work to provide this change in energy: the engines (presumably with the same force as before) and air resistance.
Wnet = We + War = 2.44 x 1010 J
Just as we presumed the engine force was always in the direction of motion, we can presume that the air resistance, like friction, is always OPPOSITE the direction of motion. So we can write:
Wnet = Fed - Fard = 2.44 x 1010 J
where d = 1.8 x 105 m, as calculated at the beginning of this part. Taking Fe from above, we have:
(Fe - Far)d = 2.44 x 1010 J
(1.71 x 105 N - Far)(1.8 x 105 m) = 2.44 x 1010 J
which leads to
Far = 35,444 N or 0.035 kN
I will check my work later to make sure I did not make a math error, but I hope this helps! And if there are any problems, or if you have any questions about this, please let me know.
