pbuk said:
1. Consider the amount of work done in ascending the hill: the heavy car's engine will do more work and therefore use more fuel.
Agreed.
pbuk said:
2. When descending the hill the light car doesn't use any fuel. In order to make up for the additional fuel it used on the way up the heavy car would have to use less than no fuel, which is not possible.
Not true. with no energy input or output, the car will necessarily have increased its velocity at the bottom of the hill. That is "stored fuel" to do extra work. You cannot neglect that.
pbuk said:
3. Also the constraint that both cars maintain the same constant speed throughout is violated: with no drag both cars will accelerate downhill.
Agreed. But the velocity doesn't matter: if a vehicle goes from height ##h_i## to ##h_f##, and then goes back to ##h_i##, considering only that point, the net energy will always be zero.
pbuk said:
4. If instead the cars travel the downhill leg first, then neither will use any fuel, however the constant speed constraint will still be violated.
Agreed.
pbuk said:
5. If instead we consider an electric vehicle with 100% efficiency, and also 100% efficient regenerative braking and make the same drag-less assumption then by conservation of energy both vehicles use no net energy in the round trip, and they can fulfill the constant speed constraint.
This proves the point made earlier: You restrain the acceleration while going downhill and instead of increasing the kinetic energy of the vehicle, you store it in some battery, spring, or rotating mass and use it to do extra work later on (doesn't have to be going uphill).
pbuk said:
6. If we add air resistance (the same for both cars) as in the OP things get more interesting.
No, it doesn't. It is just adding a different problem completely independent of the previous one.
pbuk said:
7. On the uphill leg the cars' engines will do work equal to ## mgh + f_ds ## where ## f_d ## is the drag force (which is constant because the speed is constant) and ## s ## is the distance travelled: the heavier car does ## \Delta_m gh ## more work where ## \Delta_m ## is the difference in the mass of the cars.
8. On the downhill leg gravity will do work at the rate of ## mg\frac{dh}{dt} ##, working against air resistance at ## f_d \frac{ds}{dt} ##, and in order to maintain speed each cars engine must work at a rate of ## f_d \frac{ds}{dt} - mg\frac{dh}{dt} ##. Now the heavier car's engine works at a rate ## \Delta_m g\frac{dh}{dt} ## less than the lighter car and in the total descent does ## \Delta_m gh ## less work.
Although all of this is true, it is actually hiding what is really going on.
This is a case of superposition of two independent problems:
- The car will go uphill and downhill and, assuming both initial and final levels are the same, this results in no energy requirement;
- There is a drag force ##F## acting on a path of length ##s## requiring an energy input of ##Fs##.
The total energy required is the sum of both: ##0+ Fs = Fs##. In fact, it doesn't matter:
- whether ##F## is constant or not (i.e. velocity is constant);
- whether the resistance force is a drag force, rolling resistance, coming from a big magnet pulling the car back, or the sum of all of these forces;
The total energy required will always be ##\int_{s_1}^{s_2} F_{net}ds##, just like assuming there is no uphill or downhill.
pbuk said:
9. Because the heavier car does the same amount less work in the descent as it did more in the ascent, it uses the same amount of fuel.
Yes, it does the same amount of work, i.e. drag force times the distance traveled. Your equation in point #8 ## f_d \frac{ds}{dt} - mg\frac{dh}{dt} ## only means (assuming constant velocity):
$$f_d \frac{ds}{dt} = mg\frac{dh}{dt}$$
$$f_d\cancelto{}{\frac{ds}{dt}} = mg\frac{dh}{ds}\cancelto{}{\frac{ds}{dt}}$$
$$f_d = mg\sin\theta$$
If we assume the slopes of the hill ##\theta## are the same on both sides (which it doesn't have to be) and a constant drag force (constant velocity), then:
$$E_{up} = f_d\frac{s}{2} + mgh$$
$$E_{down} = f_d\frac{s}{2} - mgh$$
$$E_{total} = E_{up} + E_{down} = f_d s$$
Assuming the slopes are different will only create a different ratio of distance traveled uphill versus downhill, but, in the end, only the total distance traveled ##s## matters.
