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I am trying to calculate the energy/ power required to move a plate up and down in a liquid.

My approach is to calculate the forces that must be applied by the motor to push/pull the plate based on Newton's law (as a function of the velocity, weight of the plate, properties of the fluid, and so on). Then to calculate Work as the product between force and displacement.

Can I say that the outcome is the required energy? Or is this a misleading assumption, based on the fact that both work and energy have the same measuring unit?

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  • $\begingroup$ energy is something pretty physical... Work done is the mathematical formalism of it. $\endgroup$
    – Shing
    Commented Dec 27, 2016 at 7:39
  • $\begingroup$ @Shing What you say is not true. Both work and energy are physical quantities. Work is one way to transfer energy from one system to another. $\endgroup$
    – user139175
    Commented Dec 27, 2016 at 8:10
  • $\begingroup$ @yoric thanks for your comment, I did miss the word transfer; however, work done is a mathematical consequence of Newton's II law (that's what I meant by a mathematical formalism), whereas energy goes beyond Newton's mechanics to relativity (at least in special relativity). $\endgroup$
    – Shing
    Commented Dec 27, 2016 at 10:15
  • $\begingroup$ @Shing Forces are well-defined in Special Relativity as well, and hence work. What changes is the explicit relation between kinetic energy and the other quantities. To me, the definition of work looks not more artificial than the (non covariant!) identification of energy as the component 0 of a 4-vector (the 4-momentum). $\endgroup$
    – user139175
    Commented Dec 27, 2016 at 11:29

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Yes, this is basically the correct approach. Strictly speaking, some of the energy you put in will go into heating the water, i.e., the water will get hotter as you agitate it. The energy balance would be something like:

$$ \Delta E = W + C\Delta T $$

where $C$ is the specific heat and $\Delta T$ the temperature rise. However, under most conditions, the temperature rise will be small enough that it can be ignored.

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  • $\begingroup$ In other words, as long as we can ignore other ways to transfer energy (e.g. heat) or other ways in which it can be stored (e.g. thermal energy) the balance Work = Energy is correct. $\endgroup$
    – user139175
    Commented Dec 27, 2016 at 8:14
  • $\begingroup$ I am moving the plate really slowly (2 mm/s) , so I think we can ignore the generated heat. $\endgroup$
    – Pompilia
    Commented Dec 27, 2016 at 8:24
  • $\begingroup$ On the left hand side we should have the variation of the energy, shouldn't we? $\endgroup$
    – gented
    Commented Dec 27, 2016 at 8:59
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    $\begingroup$ en.wikipedia.org/wiki/Thermal_efficiency Without more information the Work = Energy is a good approximation, but should never be taken as granted. In general you should avoid saying such thing. For the rigorousness you may want to say that approximation is used. $\endgroup$
    – user3644640
    Commented Dec 27, 2016 at 11:16
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Apparently (based on other answers) it depends on what your specific teacher(s) say(s)...

But yes, work and energy are the same thing, with only connotative differences; namely that work is the amount of energy by which a system changes:

$W=\Delta E$ (minus entropy losses if you're being picky)

I'd say it's like time vs. duration, if that helps.

If you have a known/required speed at which you need to move this piston thing, then the product of force and speed (or quotient of work and time) gives you power:

$P = \frac{\Delta Energy}{Time} = Force \frac{\Delta Distance}{Time} = Force \times Speed$

$P = \frac{\Delta Energy}{Time} = \frac{(Force)(\Delta Distance)}{Time} = \frac{Work}{Time}$

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  • $\begingroup$ @ Andrew Tofelt Yes, I know the speed (variable in time) at which I need to move the plate. $\endgroup$
    – Pompilia
    Commented Dec 27, 2016 at 17:55
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Energy is a property of a body or system. Things have energy.

Work is a property of an interaction and represents a transfer of energy. Things do not have work, they do work (or have work done upon them).1

This distinction may seem niggling to you---it certainly did to me when I was learning---but it is useful in making communication between humans clear and making the words match the math. It is worth your time to school yourself in keeping the two ideas (energy and transfer of energy) distinct in your head.

If the circumstance are well enough defined you can equate work done with a change in energy (indeed, that is exactly what we're doing in the work-energy theorem), and if you know the time-scale of the process then you can use that to compute the power needed.

1 Thinking ahead a bit we could mention here that heat is another transfer and objects never have heat in physics. What the chemists say is a different story.

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  • $\begingroup$ Ok, I have changed the title of the posting, because this is actually the question I wanted to ask. I know the distinction between doing work and having energy. $\endgroup$
    – Pompilia
    Commented Dec 28, 2016 at 5:34
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Can I say that work = energy?

No. Because it isn't. Work is the transfer of energy.

I am trying to calculate the energy/ power required to push a plate up and down in a liquid.

Energy is not the same thing as power. Power is the rate of doing work.

My approach is to calculate the forces that must be applied by the motor to push/pull the plate based on newton's law (as a function of the velocity, weight of the plate, properties of the fluid, and so on). Then to calculate Work as the product between force and displacement.

Yes, work is force x distance. No problem with that.

Can I say that the outcome is the required energy? Or is this a misleading assumption, based on the fact that both work and energy have the same measuring unit?

The outcome is an energy transfer from say a battery to the motor thence to the water. In this case you heat the water and the apparatus.

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  • $\begingroup$ I did not say energy is the same thing as power. I said I want to express my results as energy or power. $\endgroup$
    – Pompilia
    Commented Dec 27, 2016 at 13:18
  • $\begingroup$ Ok. If they are not equivalent, then how do I calculate the energy or power necessary to move the plate up and down? $\endgroup$
    – Pompilia
    Commented Dec 27, 2016 at 13:25
  • $\begingroup$ Pompilia : you need to gather your variables. Like the diameter of the plate, the distance you're going to move it, the time you're going to move it in, and the density and viscosity of the liquid. Is the apparatus confined within a tube? Perhaps you should ask another question about all this. $\endgroup$
    – John Duffield
    Commented Dec 27, 2016 at 13:28
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    $\begingroup$ I have gathered my variables. Ok, it seems you can only tell me what should NOT be, but not what SHOULD be. Let's let others to answer, then. $\endgroup$
    – Pompilia
    Commented Dec 27, 2016 at 13:30

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