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We suppose, however, that there is such a thing—a reversible machine—which lowers one unit of weight (a pound or any other unit) by one unit of distance, and at the same time lifts a three-unit weight. Call this reversible machine, Machine A. Suppose this particular reversible machine lifts the three-unit weight a distance X. Then suppose we have another machine, Machine B, which is not necessarily reversible, which also lowers a unit weight a unit distance, but which lifts three units a distance Y. We can now prove that Y is not higher than X; that is, it is impossible to build a machine that will lift a weight any higher than it will be lifted by a reversible machine. Let us see why. Let us suppose that Y were higher than X. We take a one-unit weight and lower it one unit height with Machine B, and that lifts the three-unit weight up a distance Y. Then we could lower the weight from Y to X, obtaining free power, and use the reversible Machine A, running backwards, to lower the three-unit weight a distance X and lift the one-unit weight by one unit height. This will put the one-unit weight back where it was before, and leave both machines ready to be used again! We would therefore have perpetual motion if Y were higher than X, which we assumed was impossible. With those assumptions, we thus deduce that Y is not higher than X, so that of all machines that can be designed, the reversible machine is the best.

Why is that?

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    $\begingroup$ Well, Feynman explains why right there. What part of his logic do you have trouble with? $\endgroup$ – knzhou Sep 24 '18 at 13:09
  • $\begingroup$ My question is.. He said that "obtaining free power" is that why that's not possible? And again he said in the last line of his argument that "and leave the both machines to be used again!" how this thing means that we would have perpetual motion?? $\endgroup$ – Anand Sahegal Sep 24 '18 at 13:13
  • $\begingroup$ It's about "conversation of energy". You would end up with more energy then you started, which violates "conversation of energy" $\endgroup$ – Hilmar Sep 24 '18 at 14:35
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It takes some pondering, but it does make sense. For clarity, I'll use kg and m just to illustrate the point. (Note that all I am doing in this answer is giving the above text in my own words, in the hope that you might see why it works from someone else's mind, though I do not claim to be mightier than the great Feynman! :P)

Machine B drops the 1kg weight 1m, and pulls up the 3kg weight by $Y$ (in m).

Now the 3kg weight is dropped from $Y$ meters above the ground to $X$ meters above the ground - i.e. if $Y>X$, then the 3kg weight has dropped by $Y-X$, which means this is giving us some energy/work (we could in principle attach this to another pulley system to lift up something else) - this is what the "obtaining free power" is referring to.

Now the 3kg weight is dropped from $X$ to 0, and using machine A, the 1kg weight is pulled back up to its starting position. This is because Machine A is reversible - this point is key to the argument.

Now think about what we've just done - we've gotten back to exactly how we started, and have also extracted some energy! This is a no-no, because it means we could theoretically do this forever and generate infinite energy!

Note that if Machine A were irreversible, then it would (potentially) be ok, because bringing the 3kg weight back from $X$ to 0 would not bring the 1kg block back to its starting position, and we could not keep repeating the above for ever without inputting some external energy. In that case, it might be allowed that $Y>X$.

What we conclude from this thought experiment is that reversible machines are the most efficient, and further to that, all reversible machines must have the same efficiency.

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  • $\begingroup$ Thank you.. I think now the argument is quite clear for me.. I just wanted to ask one thing "This is a no-no, because it means we could theoretically do this forever and generate infinite energy!" is the energy we got from the process will be sufficient to repeat the process again?? $\endgroup$ – Anand Sahegal Sep 24 '18 at 14:27
  • $\begingroup$ Yes if Machine A is reversible, because this takes the system back to its initial conditions, and if it was able to move the 3kg weight up by $Y$ in the first place, reproducing these initial conditions should allow this to happen again. If Machine A is not reversible, then no, the 1kg weight would be put back to a different place, and then acting Machine B again would not be able to get the 3kg weight up to $Y$. $\endgroup$ – Garf Sep 24 '18 at 14:30

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