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Starting from the experience of joules we know that the mass before falling has a potential energy $mgH$ where $H$ is height from floor. Once we reach the height $0$ we observe that the kinetic energy of the mass is not equal to the initial potential energy. Joule deduced that the energy was turned into something else.

Now following this reasoning should we not write that $\Delta U = mgH-\frac{1}2mv_0 ^ 2$?

$\ v_0$ is velocity when $H=0$;

Why do we write $\Delta U = W=mgH$?

Obviously I'm considering a specific case where $Q = 0$.

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If you read the details of the Joule experiment you will find that the weight fell at essentially constant velocity. That would mean no change in kinetic energy of the mass from the beginning to the end of the fall. The increase in kinetic energy to get the weight started equaled the decrease in kinetic energy to bring it to a stop. The only change in energy of the falling mass was a change in potential energy. Therefore $W=mgh$.

However Kinetic energy is not equal to 0 so 𝑊≠𝑚𝑔𝐻

It is the change in kinetic energy that is equal to zero, not the kinetic energy itself. If the change in kinetic energy from start to finish is zero, there is no gain in kinetic energy that needs to be accounted for. Only the change in potential energy needs to be accounted for. And that is accounted for by the paddle wheel work.

You are probably aware that if you lift a mass off the ground starting at rest, raise it at constant velocity and bring it to rest at a height $h$ there is no change in kinetic energy. Therefore the positive work you did in raising the mass is $mgh$. At the same time gravity did negative work taking the same energy you gave the mass and storing it as gravitational potential energy. The net work done on the mass is zero by the work energy principle which states that "the net work done on an object equals its change in kinetic energy".

The Joule experiment reverses this process. Gravity does positive work in lowering the mass but does so at constant velocity so there is no change in kinetic energy, only a change in gravitational potential energy. Therefore the work done by gravity equals the paddle wheel work

I think you may be confusing what is going on in the Joule experiment with what would happen if the mass were in free fall. In free fall the mass accelerates increasing its velocity on the way down and therefore increases its kinetic energy. According to the description of the Joule experiment, however, the velocity of the descent of the weight was controlled to be constant. Therefore there was no net change in the kinetic energy of the mass.

Hope this helps

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  • $\begingroup$ However Kinetic energy is not equal to $0$ so $W \neq mgH$ $\endgroup$
    – ABC
    Aug 18, 2019 at 13:34
  • $\begingroup$ I said the CHANGE in kinetic energy is zero not the kinetic energy is zero. You need to understand the work energy principle. I will add that to my answer. Stand by. $\endgroup$
    – Bob D
    Aug 18, 2019 at 14:57
  • $\begingroup$ First of all, thank you. I found what I wrote on the internet, so I suppose what I read is wrong. So let's see if I understand correctly. The mass has potential energy since it is at a height H. I proceed by making a calculation and it is exactly mgH by definition of work. After that I allow the descent and I should observe a job equal to the difference of the kinetic energy, but in this case it is 0 !! Being zero, Joule deduces that it has been transformed into internal energy. It's right? $\endgroup$
    – ABC
    Aug 18, 2019 at 15:37
  • $\begingroup$ @ABC Essentially right. I would let Joule speak for himself (via his report) on what he deduced. But in summary he showed that the increase in the water temperature due to the mechanical work done by the mass, mgh, is equivalent to heat causing the same temperature increase. In other words, there is a mechanical work equivalent to heat. As there is no change in the kinetic energy of the mass, it is only the mechanical work mgh, that has been transformed into internal energy. $\endgroup$
    – Bob D
    Aug 18, 2019 at 15:54
  • $\begingroup$ I have a little doubt, the initial kinetic energy is 0, as soon as I set in motion the mechanism takes on value constant until it touches the ground ... so I have to consider the kinetic energy not initial but shortly after the start? $\endgroup$
    – ABC
    Aug 18, 2019 at 16:21

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