# Homework Help: Fallacy in Work-Energy theorem?

1. Dec 13, 2005

### rushil

Fallacy in Work-Energy theorem??

Consider a block moving on a floor which has some friction. It is given an initial velocity of 10 m/s and it comes to rest after some time due to friction. Since friction slows the block down, heat is generated, some of which goes into raising the temperature of the block.

Now consider the same above situation in a reference frame of a man moving with a constant velocity of 5 m/s. In this reference frame, the change in kinetic energy is zero... So how do we get from energy conservation that the temperature of the block rises???

2. Dec 13, 2005

### Danger

No offense intended, lad; it might be a language gap. Can you possibly rephrase your question in a way that makes sense?

3. Dec 13, 2005

### rushil

Rather tha rephrasing the question - can you please state what exactly you did not understand!! Thanks!

4. Dec 13, 2005

### Danger

Not necessary now. I read this thread before your other one. Once I saw it, I knew what you meant.
What threw me off is that you referenced velocities for both the block and the man, but gave only vector figures. There was therefore no way to know in which direction they were moving in relation to each other. That sort of made the whole question ambiguous to me. Again, I meant no offense by my response.

5. Dec 13, 2005

### rushil

Forget velocities - the block and person are all travelling in the same line , initially , the block and person have earth frame velocities in the same direction .... hope its OK now!!

6. Dec 13, 2005

### Danger

It's cool, man. As I said, your second post, and Mr. Monkey's excellent response, cleared up what I missed the first time around.

7. Dec 14, 2005

### pervect

Staff Emeritus
You have to include the energy of the Earth.

The change in energy of the Earth is .5*me*ve^2 - .5*me*(ve-p/me)^2, where ve is the velocity of the Earth, me is the math of the Earth, p is the amount of momentum given to the block by the Earth. Thus the Earth's velocity changes by p/me.

You can see that there is a term proportional to p*ve and another term proportioanl to .5*p^2/me in the expression for the change in energy of the Earth.

Because me is very large, the second term is always negligible, but the first term is not negligible if ve is not zero.