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Suppose that the air resistance a car encounters is independent of its speed. When the car travels at $15\mathrm{~m/s}$, its engine delivers $20\mathrm{~hp}$ to its wheels. (a) What is the power delivered to the wheels when the car travels at $30 \mathrm{~m/s}$? (b) How much energy does the car use in covering $10\mathrm{~km}$ at $15\mathrm{~m/s}$? At $30\mathrm{~m/s}$? Assume that the engine is $25\text{%}$ efficient. (c) Answer the same questions if the force of air resistance is proportional to the speed of the automobile. (d) What do these results, plus your experience with gasoline consumption, tell you about air resistance?

The answer key says

a. $40\mathrm{~hp}$; b. $39.8\mathrm{~MJ}$, independent of speed; c. $80\mathrm{~hp}$, $79.6\mathrm{~MJ}$ at $30\mathrm{~m/s}$; d. If air resistance is proportional to speed, the car gets about $22\mathrm{~mpg}$ at $34\mathrm{~mph}$ and half that at twice the speed, closer to actual driving experience.

I understand parts (a) and (b). But in part (c), why does the power delivered to the wheels double again? Is the problem implying that all of the power delivered to the wheels is used to overcome air resistance?

It seems to me like some share of it must be used to overcome internal friction, friction between the wheels and road, and other inefficiencies—quantities that would remain constant even if the air resistance increased.

If the answer to the question in bold is simply yes, then I'd be happy to accept a one-word answer. But I'm wondering if there's something more subtle going on, e.g. if those other forces also have a linear relationship to air resistance.

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  • $\begingroup$ This post is about a fan blowing air, but it applies to a car driving through air too. physics.stackexchange.com/q/536074/37364. Internal friction, particularly in the engine dominates at low speed. I have heard that around 35 mph, air becomes more important. Air dominates at high speed.But I don't have a model of engine friction. $\endgroup$
    – mmesser314
    Commented May 8, 2020 at 3:14
  • $\begingroup$ Thanks for the link. So—yes? $\endgroup$
    – Max
    Commented May 8, 2020 at 3:16
  • $\begingroup$ The relationship is not linear. It is cubic. Yes and no. At low speed it is a better approximation that all of the power overcomes engine friction. The faster you go, the better the approximation that is all goes to air resistance. Fluid dynamics is very complex. It full of rules of thumb that simplify life, even if they are not perfect. $\endgroup$
    – mmesser314
    Commented May 8, 2020 at 3:22
  • $\begingroup$ I mean in the context of the question posed. In part (c), doubling the velocity doubles the air resistance, which doubles the power consumption (compared to the 30 m/s scenario where the velocity increases without doubling the air resistance)—a linear relationship. From your comment, I deduce that the answer to "Is the problem implying that all of the power delivered to the wheels is used to overcome air resistance?" is yes. Is that right? $\endgroup$
    – Max
    Commented May 8, 2020 at 5:35
  • $\begingroup$ In part c, the idea that doubling the velocity doubles the air resistance is wrong. The link shows why air resistance goes up 8 times. As I said above, to does all the power used to overcome air resistance: that is a better approximation at high speeds than low speeds. $\endgroup$
    – mmesser314
    Commented May 8, 2020 at 13:47

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