# Work and Energy Confusion

1. Nov 4, 2014

### Elliot Webb

I apologise for what is probably merely stupidity on my part, but has been vexing me for rather a while, and no one I ask can provide any insight.

A body undergoes two accelerations. The first is from 0 to 10 m/s, the second from 10 to 20 m/s. For simplicity's sake I will say F= 1N, a= 1 m/s2 and m = 1kg, for both of the accelerations.

So in both cases, it takes 10 seconds for the particle to accelerate, and so if it were powered by some kind of engine, the time it is running would also be 10 seconds. As the force it produces is constant, the amount of fuel burnt in each case would logically be the same. But apparently it's not.

Here is the problem:
The distance travelled in the first instance is 50m, with u=0, v=10, a=1. As work done= Fs, the energy transferred is therefore 50J. So 50 joules worth of fuel is burnt.

For the second acceleration, u=10 and v=20, so the distance travelled is 150m. Hence the energy transferred the second time is 150J.

So despite the engine providing the same amount of force for the same time, it has used three times as much fuel. Unless it knows how fast it is going (with relation to what, anyway?), how could it take more energy to accelerate the second time?

Thank you for you patience and (hopefully) your help.

2. Nov 4, 2014

### CWatters

Nope that's not logical. A fridge magnet applies a constant force but consumes no energy.

Work = force * distance

Divide both sides by time and you get..

work/time = force * distance/time

or

Power = force * velocity

It's logical that it takes more power to go faster even with a constant force.

3. Nov 4, 2014

### Elliot Webb

If it takes more power, how does the engine 'know' how fast it is going in order to be less efficient? If you were to just leave it running at the same power, assuming no external resistance, the force produced would gradually decrease as the body accelerates. I don't see how this could be possible. If it is, then what is the velocity in relation to? If something is in orbit, the thrusters etc. will have exactly the same effect as if the object was stationary on earth, and yet the changes in kinetic energy would be massively different - with the same amount of chemical energy being used up (fuel).

4. Nov 4, 2014

### Staff: Mentor

None of this has anything to do with efficiency.

For an internal combustion engine, you can consider the force applied and fuel used to be roughly constant for every piston stroke with the gas pedal in the same position. But if you are moving faster, the engine is spinning faster, so you have more piston strokes in the same amount of time. So more power and more fuel used.

5. Nov 4, 2014

### Elliot Webb

Until you change gear, and then the engine slows down again.

6. Nov 4, 2014

### Staff: Mentor

True. And what else changes when you change gears...?

7. Nov 4, 2014

### A.T.

I think it's less a matter of logic, than just of how those quantities are defined in physics vs. the intuitive ideas based on common language usage of the same words.

8. Nov 5, 2014

### Elliot Webb

I do understand how these are defined, but that is what has caused the problem. I don't see how there can be a massive increase in energy for something moving in orbit (for instance), compared to something that is stationary. The KE increase would be thousands if times greater, even though the same amount of fuel would be used up. The energy stored in the fuel can't change, and so where does this come from?

Also, can anyone direct me to a proof of W=Fs that doesn't start with the kinetic energy formula - thanks

9. Nov 5, 2014

### A.T.

Did you compare the change in KE of the fuel when it's expelled by the rocket in both cases?

10. Nov 5, 2014

### Lsos

It might not be "logical" at first glance, but when you start to study the everyday things around you you will see that this is how it works. For example, if you jump out of of a 2nd story window, you will NOT hit the ground 2x faster than from a 1st story window (0 story being the ground floor).
It's also why it takes 8x more power/ and hence 8x more fuel to drive 200 mph as opposed to 100 mph. Fuel efficiency drops off really fast above highway speeds, much faster than air resistance increases.
It's also why military jets produce many tens of thousands, if not hundreds of thousands of horsepower, despite only being able to go like 6-10x faster than a car (at sea level).
It's also why crashing into a wall at 20 mph is 4x more destructive than at 10 mph.
The examples are all around you.

How does an object KNOW how fast it is moving? It knows because it has to push against something in order to provide a force. A car has to push against the ground. The faster the car moves, the faster the ground recedes, and you need to change gears just to keep up, trading force for distance.

I recommend you don't think about rockets until you understand the above, because a rocket is a special case and can get confusing. But the jist of it is this: the difference in a rocket is that it pushes against its own exhaust. It essentially brings the ground with it. So yes, if a rocket travels fast enough it can indeed gain more energy than what is stored chemically in its fuel, but that energy didn't come from nowhere. It came from the kinetic energy that was put into the fuel from the beginning of the flight. After all, an orbiting rocket's fuel has ALOT of kinetic energy in it, much more in fact than its chemical energy. That energy was "invested" into the fuel from liftoff, and that energy gets returned at high speeds. This is called the Oberth Effect btw.

Last edited: Nov 5, 2014
11. Nov 5, 2014

### Elliot Webb

Yes, and it makes the problem even worse. With the first case, say for each of the accelerations 10 kg of fuel is expelled at 1 m/s for 10 seconds. The KE if the fuel is 80 J, so the total KE of the rocket and fuel is 130 J. After the second acceleration, the total KE is much more - 1180 J. If I were to calculate this for real rockets travelling at thousands of km/h then tiny accelerations would consume immense amounts of energy- and the exhaust doesn't help.

12. Nov 5, 2014

### jbriggs444

Go ahead and calculate it. You will find that the apparent discrepancy is covered by a careful accounting for the energy in the exhaust stream.

13. Nov 5, 2014

### Lsos

Again, this is called the Oberth Effect and is a special case for rockets.

But not really, because you must keep in mind that kinetic energy depends relative to what you calculate it. We usually calculate kinetic energy relative to some stationary object (the earth), but a walk in the park could literally be millions of horsepower when compared to some receding globular cluster on the other side of the universe. That energy is not from nowhere though, it was put into you and the earth and the globular cluster at some distant point in time, by whatever caused them to fly apart from each other in the first place.

14. Nov 5, 2014

### A.T.

If the same rocket burns for the same time, then the fuel used will be obviously the same.

15. Nov 5, 2014

### A.T.

How did you compute that?

16. Nov 5, 2014

### Lsos

You're right, I removed that entire sentence because I thought OP meant something completely different.

17. Nov 5, 2014

### Staff: Mentor

Much of that is wrong. Air resistance is a square function of speed and so if you double the speed you get 4x the air resistance. Then combining the air resistance and higher speed, 2x4=8 times the power. And none of that has anything to do with efficiency. If you meant fuel economy, which is sort of a measure of efficiency, it is still wrong since that is in terms of distance, not time. So the 2x speed factor goes away and the drop in fuel economy is due entirely to drag and other frictions... all still assuming no change in thermodynamic efficiency of the engine. In other words, it takes 8x as much power but only 4x as much fuel per mile.

Again, let's keep efficiency out of this. The ideal case of no drag and ignoring engine efficiency issues and focuses on the acceleration from 0-100 and 100-200. Adding efficiency is a separate, confounding factor that won't help answer the original question.

Bottom line is that if you want acceleration to be the same at 200 mph as at 100 you need twice the power because you are covering twice the distance in the same time. And that manifests as the engine spinning twice as fast. Simple.

Or alternately if you want to keep the rpm constant by changing gears.... what else must change? (Question I previously asked of the OP that went unanswered). Again, the relationship here between force, rpm, speed, power and acceleration is simple proportions. If one changes another must change by the same amount. It isn't a trick question and no detailed analysis is required.

Last edited: Nov 5, 2014
18. Nov 5, 2014

### CWatters

This has nothing to do with efficiency.

The engine doesn't know anything. In order to accelerate from 10 to 20M/S you (the driver of the car) will find they have to increase the amount of fuel fed to the engine compared to 0 to 10M/S. eg If you don't step on the gas it won't accelerate from 10 to 20M/S as fast as it did from 0 to 10 M/S.

19. Nov 5, 2014

### Elliot Webb

Thanks for pointing out the Oberth Effect - I'd never heard of it and didn't consider that the fuel itself has kinetic energy. At some point I'll do some Maths and see if it satisfies the problem. I apologise for mentioning efficiency - I didn't mean it in the traditional sense, I just meant that for the same amount of fuel being used, less force is produced, which to me seems less efficient, but maybe I used the wrong word. However, you keep asking that 'what else changes when you change gear?' Although the wheels may be at a different speed, as long as the engine RPM is the same, surely the same amount of fuel is being consumed in a certain time? It can be seen quite easily that as long as the engine RPM remains the same, the rate of fuel consumption stays constant, irrespective of the speed of the car - otherwise maintaining a high speed would be next to impossible.

I thank you all for helping, but I'm afraid I'm still not satisfied. Also, has anyone got a proof of the Work-Energy principle that doesn't begin with 'assume KE=1/2mv^2'?

20. Nov 5, 2014

### jbriggs444

If engine RPM remains the same and the vehicle goes faster, then you must have shifted into a higher gear. Even if air and rolling resistance remain constant (a dubious assumption), the engine will still have to push harder to make up for the higher gear ratio. You will have to press down harder on the gas pedal to open up the throttle. Fuel consumption will increase accordingly.