Regarding the work done against air resistance

[vodkasoup]

Homework Statement

I am confused by the answer to a question given in an exam past paper. The question asks the work done by an aircraft engine over a period of time and a calculation of air resistance. I have calculated the power of the engine, the work done over a certain time period, and the gain in potential energy. However, I have a question regarding the final calculation of the work done against air resistance.

Relevant Equations

P = Fv
U = mgh
Work = Fd

Here is the question:

I have correctly calculated the power produced by the thrust force (P = Force x Velocity = 9.0 × 10⁴W) , the work done by the thrust force over 3 minutes (W = Power x Time = 9.0 × 10⁴ × 3.0 × 60 = 1.6 × 10⁷ J) , and the gain in potential energy over this period (mgh = 1200 × 9.81 × 3.3 × 3.0 × 60 = = 7.0 × 10⁶ J ). The confusion sets in when I try to calculate the work done against air resistance during this time.

I know that the energy 'lost' to air resistance is equal to the total work done minus the energy of the aircraft, i.e., the energy that is 'missing' from the total work done. The marking scheme states that this is equal to the work done minus the gain in potential energy, which comes out as (1.6 × 10⁷ J - 7.0 × 10⁶ J = 9.0 × 10⁶ J .

My question is, why is the kinetic energy of the aircraft not included in this calculation? Surely there is work being done by the engine to produce the kinetic energy which drives the aircraft forward? Should the energy 'lost' to air resistance not then be the total work done minus the sum of both the potential and kinetic energies?

Is it to do with the fact that only the potential energy changes, and not the kinetic energy? Is the 3.0m/s gain in altitude therefore only the vertical component of the 45m/s velocity?

Many thanks for all replies.

 

[etotheipi]

My question is, why is the kinetic energy of the aircraft not included in this calculation? Surely there is work being done by the engine to produce the kinetic energy which drives the aircraft forward?

No, if the velocity is constant (as we are told) then the kinetic energy is constant ($KE = \frac{1}{2}mv^{2}$!). The work energy principle when expanded to include potential energies goes something like

$W_{non-conservative} + KE_{initial} + PE_{initial} = KE_{final} + PE_{final}$

Evidently the thrust force does non-conservative work, and air resistance is also going to do some negative work. Since the initial and final KE is the same, we can cancel these terms. Essentially you are left with the total work done by non-conservative forces (thrust and air resistance) equals the change in gravitational potential energy.

And as a final note, I find it easier to translate 'work done against something' to 'the negative of the work done by that thing'. I don't particularly like the phrasing work done against, it seems like an unnecessary and confusing way of avoiding negative signs.

 

[kuruman]

A more intuitive way to think about this is in terms of energy per unit time accounting. The engine input is $P_{in}=F_{thrust}v$ units of energy per unit time. That is constant. What does this energy input buy you? It buys you increased potential energy, $P_{height}=mg\frac{\Delta h}{\Delta t}$ and the privilege to move against air resistance, call that $Q$, which is what you have to find. It does not buy you kinetic energy because the speed is constant. All the energy input per unit time must be spent since energy is neither created nor destroyed. Therefore
$$ P_{in}=P_{height}+Q~\rightarrow~F_{thrust}v=mg\frac{\Delta h}{\Delta t}+Q$$The rest is algebra and substitution. Note that this equation gives you the mechanical work lost to air resistance per unit time. To find a total energy lost to air resistance, you will need a time interval to multiply it by. Yes, as you can see from the above equation, $\frac{\Delta h}{\Delta t}=3.3~\mathrm{m/s}$ is the vertical component of the velocity.