# Energy, once again weird unclear thing

1. Feb 14, 2010

hi.. once again my thoughts bumped into weird energy question, yet, very basic......

so say there is car, which mass is 1kg (yes, very realistic indeed), accelerating from 0 to 2 m/s.
to viewer on the ground the energy at start was 0 and in the end 2J, the work done by the engine to the car is 2J.
but now, and i'm quite sure most of you get the problem as soon as i say the next sentence, say i'm walking against the car direction in 1m/s speed. so for me, the initial speed is 1m/s and the final speed is 3m/s, as you can calc yourself, for me the work the engine did is now 4J.

we can get the same results if we use the W=F*S definition.

am i missing something? or does work depends on the viewer?
if so, does it not related to very practical things, such as the amount of fuel was burned and so on?

thank you very much indeed :D

2. Feb 14, 2010

### Staff: Mentor

Yes, energy is frame variant or coordinate dependent.

It does relate to the amount of fuel burned in any given frame. Energy is frame variant, but it is also conserved.

Go ahead and work out the full problem in both frames taking into account the conservation of momentum and the KE of the earth. For this example it is ok to use any arbitrary large mass for the earth and to consider linear momentum rather than angular momentum.

3. Feb 14, 2010

i did NOT ask about energy, but about work. i know obviously that energy is frame variant.. but work as well?

i really have no idea what the earth movement, mass, and even momentum, have to do with that......
and what do you mean by "work out the full problem"? what exactly should i do?

4. Feb 14, 2010

### Staff: Mentor

Yes.

Assume some large mass of the earth. Start with the earth and car initially at rest and use conservation of momentum to determine the the initial and final velocites of the earth and the car. Use those to determine the change in the energy of the system. Then do the same except starting with the earth and the car initially at 1 m/s.

5. Feb 14, 2010

### Staff: Mentor

A couple of things not noted yet:
Work isn't force times speed, it is force times distance.
Yes, what you are missing is that the car isn't doing any work against you, so that frame of reference doesn't relate to how much energy the engine expended. The car is sitting on the ground, so the work done by the engine can only be calculated in that frame.

6. Feb 14, 2010

### cesiumfrog

"Speed"? Doesn't s conventionally stand for displacement?

What? Only?
"That frame" isn't even inertial..

7. Feb 14, 2010

### Staff: Mentor

The work done can be calculated in any frame. You will get different amounts of work done, but as long as you consider the entire system you will get conservation of energy. Here is a quick example of the kind I was encouraging the OP to work out:

Let's say that we start with the 1 kg car and a 1000 kg earth both initially at rest and the car accelerates to 2 m/s. By conservation of momentum the earth will accelerate to -0.002 m/s. The change in the KE of the car is 2 J and the change in the KE of the earth is 0.002 J, so the engine did 2.002 J work which is the change in the total KE of the system.

Now suppose both the car and the earth are initially travelling at 1 m/s and the car accelerates to 3 m/s. By conservation of momentum the earth will decelerate to 0.998 m/s. The change in the KE of the car is 4 J and the change in the KE of the earth is -1.998 J, so the engine again did 2.002 J work which is the change in the total KE of the system.

So you can calculate the work done by the engine in any frame, you just have to be careful and not neglect important parts of the system.

Last edited: Feb 14, 2010
8. Feb 14, 2010

### Staff: Mentor

Ehh, that's the engineer in me.... didn't realize that displacement is s. But from the OP it looks like he's multiplying force times speed: he doesn't say what his force is, but when he talks about the first case having a speed of 2 m/s and 2J of KE and the second case being 4m/s and 4J, that's what it looks like to me. This can't be since KE is a square function of speed.
Before I get to a full response, isn't the earth's new speed 0.998 m/s?

That little issue aside, *I don't think it is important to consider the change in earth's speed since it
earth is so much larger that the earth's change in energy is near zero. In either case, yes, you can calculate work in any frame, but at the end of the problem, you're still subtacting one from the other to find the work done by the car against the ground. You're just converting frames at the end of the the problem instead of selecting the relevant frame at the start of the problem.

[edit2] *Nevermind, I figured out what you did. I didn't realize the earth's energy change would be significant for earth (that's me not being a physicist...). Nevertheless, at the end you are subtracting the work done on earth as seen by the person from the earth done on the car as seen by the person, which gives you the answer in a car vs earth frame. You're still just converting frames at the end of the problem instead of selecting that frame from the beginning, aren't you?

Last edited: Feb 14, 2010
9. Feb 14, 2010

### Staff: Mentor

Oops, quite right. I have it correct on my notes and copied it down wrong.
Only in the frame where the earth is at rest. In other frames the earth's change in energy is significant even if the change in velocity is not. In this example the change in the earth's energy is approximately equal to the amount of work done by the engine, hardly negligible.

Last edited: Feb 14, 2010
10. Feb 14, 2010

### Staff: Mentor

See late second edit...I had to do the problem again with the earth's mass at a million kg to prove that second part to myself.

11. Feb 18, 2010

very interesting!
but didn't you tell me work is frame variant?
my common sense says it is not, and so you are showing yourself in this example..

so did i not understand you correctly, or is there still cases in which the work done by a certain force would be different in different frames?

thanks again :)

btw, i think it is better to show it with parameters..

12. Feb 18, 2010

### Staff: Mentor

I thin the disagreement DaleSpam and I were having was largely semantec in nature. The end result of the exercise is that Dale calculated the work differently, injecting different reference frames into one of the calculations, and came out with the same answer as if he only used the one frame. So no, work done does not change when viewed from a different reference frame.

13. Feb 19, 2010

### Staff: Mentor

Work done does change when viewed from a different reference frame. My point above is that energy is still conserved, even though the work is different in different reference frames. Energy is a frame-variant but conserved quantity, you just need to pay attention to the whole system.

Although my above example does not show it since energy is the timelike component of the four-momentum it "dilates" the same way that time does when you start talking about relativistic velocities.

14. Feb 19, 2010

### DaTario

You should approach this problem thinking acceleration is not frame invariant (in changing between inertial frames) ,mass, displacement and angles also not, so the product of force, displacement and cosine of the angle also won't be.

Relativity put far from this discussion, obviously...

Best wishes,

DaTario

15. Feb 19, 2010

### Staff: Mentor

Me too, I think that it is better showing general relationships. Since you prefer parameters then it is much easier:

The four-momentum is given by:
(E/c,p)
Which transforms as a four-vector.
(E'/c,p')=L.(E/c,p)

Therefore, like all four-vectors, the components are frame variant.

16. Feb 20, 2010

so can you give an example in which the work is different between one frame and another?

and i have no idea what four momentum is :P

17. Feb 20, 2010

### Staff: Mentor

The work done on the car is different in each frame.

Another simple example:
A stewardess on a jet travelling at 300 m/s ground speed holds a cart in place next to row 14 using 10 N of force for 1 s, how much work did she do on the cart in the plane's frame and in the ground's frame?
in the plane's frame: w = f.d = 10 N 0 m = 0 J
in the ground frame: w = f.d = 10 N 300 m = 3 kJ

Again, work is different in each frame, but energy would still be conserved.

18. Feb 20, 2010

### DrZoidberg

Of course the amount of fuel burnt will be the same in each reference frame. Fuel consists of atoms. Even relativistic effects can not change the number of atoms. They can change their size. They can change the speed with wich the electrons move around the nucleus etc. but not their number.
In your original example the amount of work done by the car engine will be the same in both reference frames. When you move towards the car with 1m/s it's kinetic energy is higher. However in that reference frame the ground is also moving with 1m/s. Therefore the ground will do part of the job of accelerating the car. The engine doesn't need to deliver the full 4J but only 2J.
If you look instead at a rocket the situation is similar. The amount of energy delivered by it's engine is the same in each reference frame. However the energy is distributed differently. Part of the energy goes into the rocket and part goes into the burnt fuel that is shot out the end. So if you change the reference frame and the energy put into the rocket e.g. increases then the energy put into the burnt fuel will decrease or vice versa.

Last edited: Feb 20, 2010
19. Feb 21, 2010

### Staff: Mentor

Exactly. Well said.

20. Feb 21, 2010