Fallacy in Work-Energy theorem?

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Homework Help Overview

The discussion revolves around the Work-Energy theorem and its implications in different reference frames, particularly focusing on a block moving on a frictional surface. The original poster questions how energy conservation applies when analyzing the situation from a moving reference frame, where the kinetic energy appears unchanged.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between kinetic energy changes and temperature rise due to friction. The original poster raises a question about the implications of observing the system from a different reference frame, while others seek clarification on the initial question and the direction of motion.

Discussion Status

Some participants have provided clarifications regarding the original poster's question, addressing ambiguities in the description of velocities. There is ongoing exploration of how to incorporate the Earth's energy into the discussion, with one participant suggesting that the change in energy of the Earth must be considered.

Contextual Notes

There are indications of potential confusion due to the lack of clarity in the direction of motion and the reference frames involved. Participants are attempting to reconcile these aspects while discussing the implications of the Work-Energy theorem.

rushil
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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?
 
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No offense intended, lad; it might be a language gap. Can you possibly rephrase your question in a way that makes sense? :confused:
 
Rather tha rephrasing the question - can you please state what exactly you did not understand! Thanks!
 
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. :smile:
 
Forget velocities - the block and person are all traveling in the same line , initially , the block and person have Earth frame velocities in the same direction ... hope its OK now!
 
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. :smile:
 
rushil said:
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?

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.
 

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