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If an airplane of mass $m$ is flying at a constant speed $v$, the power of the airplane is $$P = m\cdot v\cdot g $$ where $g$ is the acceleration of gravity and therefore: $$ F = m\cdot g, $$ but, if the velocity is constant, there is no net force as well as no work done. Then how can the magnitude of power be non-zero?

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  • $\begingroup$ NB: Your 2nd equation is incomplete; it's the sum of the forces that would be zero, $F=mg\neq0$ by necessity (the airplane has mass & gravity exists). $\endgroup$
    – Kyle Kanos
    Commented Apr 5, 2015 at 14:44

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The plane itself is exerting some force in order to overcome the effects of gravity.

Consider that if you turn off the engine, the plane would go crashing down. The only reason you are able to observe a constant velocity is because the plane is exerting some power to work against gravity (and other phenomena such as air-resistance etc.)

Now you may be wondering, where is all this energy exerted by the plane to overcome gravity going to? Where is the earth getting this seemingly infinite energy to pull the plane down?

Well, it may seem to you that the Earth is stationary and constantly pulling the plane, seemingly violating the First law of thermodynamics, it is crucial and absolutely imperative to note that the earth is also being attracted to the plane!

The plane is applying a force on the earth over a very tiny displacement, unobservable through the naked eye but nonetheless present.

So it is not the Earth's gravity that is doing work on the plane, but the plane that is doing work on the Earth. Remember, work is the force is applied over a displacement.

To summarize $P = mvg$ approximates the power exerted by the plane on the earth, and not the other way around.

By looking at your comments on some of the other answers, it seems to me that you have certain doubts regarding work. This is something that should help you out.

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  • $\begingroup$ what about a car that is being driven on a road at a constant velocity ?...would it's power also be non zero? because the net force on the car would be zero and work should be zero too right? $\endgroup$ Commented Apr 5, 2015 at 6:05
  • $\begingroup$ Yes, the power would be non-zero, as it would have to overcome the effects of friction, air-resistance and even retardation due to gravity. $\endgroup$ Commented Apr 5, 2015 at 6:11
  • $\begingroup$ @AkibAhmedIshan: Please observe the changes I've made to my question. It should help clear some doubts. $\endgroup$ Commented Apr 5, 2015 at 6:35
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As you say, there is a downwards force on the plane of $Mg$ where $M$ is the mass of the plane and $g$ is the acceleration due to gravity. There is also a force due to aerodynamic drag, but let's ignore that for now.

The key to understanding how the plane stays up is that force is equal to the rate of change of momentum. In this case the wings of the airplane are changing the momentum of the air they pass through. This exerts a downwards force on the air, and the equal and opposite upwards force is what keeps the plane aloft.

If we are flying in a 747 weight around 400 tonnes then the downwards force is 400,000$g$, so the wings need to change the momentum of the air by 400,000$g$ kg.m/s every second. A cubic metre of air weighs about 1.2kg (at sea level - less as the plane ascends) so one way to do this would be to accelerate a single cubic metre of air to around 3,000 km/s every second, but that's not realistic. An alternative would be to accelerate about 4 million cubic metres of air to a speed of 1m/s every second, but that's not realistic either. The actual values for the mass of air accelerated and the velocity will be somewere between these two extremes.

The actual airflow around the wings of a 747 is complicated, and the simple model I've described above can only be an approximation. But if you're willing to go along with the approximation we can derive an expression for the power. Suppose that every second the wings accelerate a mass $m$ of air to a velocity $v$. Then because force is rate of change of momentum we know that:

$$ Mg = mv \tag{1} $$

The power required is the kinetic energy of the air per second:

$$ P = \tfrac{1}{2}mv^2 \tag{2} $$

It would be interesting to use this equation to estimate the mass of ir moved and the velocity - we can do this by substituting for either $m$ or $v$ in equation (2) using equation (1). However, rather surprisingly, I cannot find figures for the power consumed by a 747 when cruising. If anyone can find these figures I can edit my question to estimate $m$ and $v$.

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  • $\begingroup$ A simple estimate for the mass of air moved is obtained by looking at the area swept by the wings per unit time - wing span times velocity. Given area and density you can estimate velocity needed for a certain impulse. Recognizing of course that at 10km air density is not 1.2 kg/m<sup>3</sup>... $\endgroup$
    – Floris
    Commented Apr 5, 2015 at 14:24
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If an aeroplane of mass m is flying at a constant speed v, the power of the aeroplane is $P=m∗v∗g$ (g being the acceleration of gravity and therefore: $F=m∗g$), but,

if the velocity is constant there is no net force as well as no work is done

The boot is on the other leg: the velocity is constant just because net work is done to keep it at the same altitude and at the same velocity.

If no work were done every second the plane would gradually loose altitude and velocity and crash onto the ground

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Assuming the plane is essentially a hover-craft that is gliding through the air without any resistance, the work must be zero over any distance it travels. The force that the plane is exerting to counter the force of gravity (and stay up in the air) is purely in the radial direction, while it is (assumed to be) moving in a direction perpendicular to it. Thus, $\mathbf{F}\cdot d\mathbf{r}$ is always zero, so it is not doing any work, and hence the power must be zero.

Of course, these are pretty tight restrictions that I've set/assumed. Air resistance is a non-conservative force in this case, and so the work exerted by the plane to overcome that force and maintain its velocity isn't zero, and so the power in that case wouldn't be zero.

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  • $\begingroup$ so if I'm dragging a block attached to a string on a frictional plane at a constant speed is the work done by me really non zero??...(I'm thinking that the force I'm exerting on the block and the frictional force is the same and opposite so the net force is 0 and work done should also be 0.Kindly clarify if there's any misconception.) $\endgroup$ Commented Apr 5, 2015 at 6:27
  • $\begingroup$ No, it is not zero in that case! Your doubt is correct. In your case, there is a force needed to push the block across the plane. The case I posited was one in which there was no friction (i.e. no 'air resistance' for the hoverboard to overcome) $\endgroup$ Commented Apr 5, 2015 at 15:39
  • $\begingroup$ What I have basically said was that a block moving across a frictionless plane at a constant velocity will not do any work, and hence no power. This is true even though there is a normal force that the plane is exerting on the block, which is always perpendicular to its velocity. $\endgroup$ Commented Apr 5, 2015 at 15:44

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