# What is behind the definitions of work and energy?

I am aware that there appear to be many similar questions on this site, but that is just because of the misleading title. I could not think of a better title that illustrates how different is this question, so if you could fix it - thank you very much.

My knowledge of physics comes down to kinematics and Newton's laws. I would like to get answers which do not deviate from this basic knowledge.

I am currently questing for exact definitions of work and energy, and what is behind these definitions . I searched in physics sites, and here in Physics in particular, but I have not found a satisfactory answer. Here is a list of the main resources I have reviewed:

Over the net:

Here on Physics:

To preface my question, I will explain the process I went through until I got my current understanding, which I want to sharpen. First, here are the main definitions of work and energy I found around the web:

Energy Is a property, or state of objects representing the ability to do work. It can be transferred to other objects, and it can be reflected in many forms, which are convertible.

Work is a transfer of energy from one object to another by applying force, and it is equal to $\vec{F_{x}}\cdot \vec{x}$ for a constant force, or to $\int_{x_{1}}^{x_{2}}\vec{F_{x}}\enspace d\vec{x}$ for a changing force.

These definitions seem circular because work is a transfer of energy, and energy is the ability to do work. To solve this problem I adopted NeuroFuzzy's approach:

Sometimes when you're stuck on things, it's helpful to look at the mathematics of what's being asserted.

A basic mathematical analysis of work definition $W=\vec{F_{x}}\cdot \vec{x}$ raises the following two points:

1. When multiplying the force or the displacement by $k$, the work is also multiplied by $k$. For example, if I apply a force of $2\vec{F}$ on an object, the work will be twice as big as if I applied a force of $\vec{F}$. Also, if a given force on an object led to $\frac{1}{2}\vec{x}$ displacement, the work it did is twice as smaller as in a case it led to $\vec{x}$ displacement.
2. A force does a positive work when:

• The object displaces.
• The force on the object is in the direction of the displacement.

A force does negative work when:

• The object displaces.
• The force on the object is in the opposite direction of the displacement.

A force does no work when:

• The object does not displace.
• No force is applied to the displacement axis.
• Both of these conditions are met.

In other words, when I apply a force on an object in certain circumstances (environment, other forces, time period, etc.) and the object's displacement is different than the displacement in the same conditions, only without the force, I could say that the force has influenced the displacement. It can have a "positive" influence (if the displacement was larger than the displacement under the same conditions without applying the force) or have a "negative" influence (if the displacement was smaller than the displacement under the same conditions, without applying the force). If I applied a force, and the displacement was equal to the displacement under the same conditions, without applying the force, then the force did not influence the displacement.

My intuitive conclusion of this analysis is that work is the degree of influence a force has on an object's displacement. If we accept my definition, and we combine it with the definitions above, we could describe work and energy as follows:

Energy Is a property, or state of objects representing the ability to apply a force that will influence an object's displacement. It can be transferred to other objects, and it can be reflected in many forms, which are convertible.

Work is the degree of influence a force has on an object's displacement. When a force does work, besides the fact that it moves the object, it also transfers to the object the ability to influence other objects' displacement by applying force.

I have a few questions about these definitions I cannot find an answer. I do not know whether the definitions are correct and the questions have answers - and then I will be grateful for answers, or the definitions are incorrect in the first place - and then I will be grateful if you could correct my definitions.

Here are the questions:

1. Does the definition of energy mean that all energy is potential energy? Because if energy describes the capability of an object to do work, doesn't it means that we are talking about the potential of the object to do work?
2. I defined energy as a property, or state of an object representing the amount of work it can do. But how could we quantify it? If, for example, a man stores within itself $100J$ of energy, does that mean he is able to apply a force of $10^{100}N$ over $10^{-98}m$? Surely a human being is not capable of applying so much force, so why we still say he stores $100J$?
3. How potential energy is reflected while being potential? Can we see a difference between a person that stores $x$ energy and a person consuming that $x$ energy? An answer I heard a lot is that the mass of the person is actually its potential energy, and therefore doing physical activities, for example, is consuming energy, and reducing the mass. But if it is correct, wouldn't we measure energy in $kg$ or mass in $Joule$?
4. Although my definitions do explain the inner nature of work and energy, I could still mathematically describe this nature in an infinite number of ways. For example, if I describe work as $2\vec{F}_{x}\cdot \vec{x}$, I would still get to the conclusion that work is the influence a force has on an object's displacement. So why is that the equation?
5. How is consuming energy in other forms than motion work? How are heat energy, or sound waves, for example, work?

Thanks.

• I think the Feynman quote is relevant here. – tfb Aug 31 '16 at 9:12
• Just a quick comment on 4., "2F⋅x". This is as correct as any other factor, including 1, because the factor is just a matter of choosing the units of force, distance and energy, which are obviously completely arbitrary and without physical significance. If you measure the distance in inches and want to obtain Joule as the result, the factor is some funny fraction. The SI system of units has been designed to eliminate many such factors in everyday equations; but it breaks down e.g. at the atomic scale (where the speed unit better be c, the charge unit the electron's etc.). – Peter - Reinstate Monica Aug 31 '16 at 16:39
• Since E=mcc, you could say that energy is any mass, that something has, that is greater than its rest mass. :) ("Potential energy" doesn't count just like a "Barber college" is not a real college. ;) ) ... Then "Work" could be defined as "throwing mass around with light, or something." :P – Brock Adams Aug 31 '16 at 16:40
• @BrockAdams Yeah... post-Newton there is no mass any longer at all, is there? "Rest mass" is only good until you meet your anti matter twin.-- But why does "potential energy" not count? Doesn't the earth-object system gain mass when you lift the object? – Peter - Reinstate Monica Aug 31 '16 at 16:43
• @PeterA.Schneider, I never was clear on what effect a gravity well had on an object's total mass. If you lift an object some distant off, say, a planet's surface (but otherwise stationary and still in the gravity well); then the system should gain some mass/energy, i think. Just not sure about the object's total mass... – Brock Adams Aug 31 '16 at 16:49

Don't be surprised that physics has a lot of definitions that are circular. Ultimately, we are just describing the universe.

Work and energy have been defined in a certain way in newtonian physics to explain a kinematic model of reality. This is a model, not reality - you will find no such thing in reality. However, in many scenarios, it is close enough to reality to be useful.

For example, let's say that a human has a 10% efficiency at converting food to mechanical work. So if you spend 1000 kJ of food energy to press against a wall, are you doing 1000 kJ of work, or 100 kJ of work, or 0 kJ of work?

In strict mechanical sense, you did no work whatsoever, and all of the energy you used was wasted as heat. If you instead used this energy to push a locomotive, you would have wasted "only" 900 kJ of the energy as heat, with 100 kJ being work. But the locomotive has its own friction, and it wil stop eventually, wasting all the energy as heat again. And overall, you did expend all those 1000 kJ of food energy that is never coming back.

All of those are simplifications. Kinematics is concerned with things that move. Using models is all about understanding the limits of such models. You're trying to explain thermodynamics using kinematics - this is actually quite possible (e.g. the kinematic theory of heat), but not quite as simple as you make it. Let's look at the fire example. You say there is no displacement, and therefore no work. Now, within the usual context kinematics is used, you are entirely correct - all of that energy is wasted, and you should have used it to drive a piston or something to change it to useful work.

Make a clear note here: what is useful work is entirely a human concept - it's all 100% relevant only within the context of your goals; if you used that "waste" to heat your house, it would have been useful work. It so happens that if you look closer, you'll see that the heat from the fire does produce movement. Individual molecules forming the wood wiggle more and more, some of them breaking free and reforming, and rising with the hot air away from the fire, while also drawing in colder air from the surroundings to feed the fire further. There's a lot of displacement - individual molecules accelerate and slow down, move and bounce around... But make no mistake, the fact that kinematics can satisfactorily explain a huge part of thermodynamics is just a bonus - nobody claimed that kinematics explains 100% of the universe. It was a model to explain how macroscopic objects move in everyday scenarios. It didn't try to explain fire.

For your specific questions, you really shouldn't ask multiple questions in one question. It gets very messy. But to address them quickly:

1. There is no potential energy in the kinematic model. The concept is defined for bound states, which do not really exist as a concept in kinematics. In other models, you might see that there's a difference between, say, potential energy and kinetic energy - no such thing really exists in reality. You need to understand the context of the model.
2. In a perfectly kinematic world, this is 100% correct. However, as noted before, kinematics isn't a 100% accurate description of reality, and there are other considerations that apply, such as the fact that humans have limited work rate, limited ability to apply force, and the materials we are built of aren't infinitely tough, perfectly inflexible and don't exist in perfect isolation from all the outside (and inside) effects. In real world applications of models, these differences are usually eliminated through understanding the limits of given models, and using various "fixup" constants - and if that isn't good enough, picking (or making) a better model.
3. You're mixing up many different models at different levels of abstraction and of different scope so confusion is inevitable. Within the simplified context of kinetics, there is no concept of "potential energy". You simply have energy that can be used to do work, and that's it; it doesn't care about how that energy is used to do work, about the efficiency of doing so etc. In another context, it might be very useful to think of energy and mass as being the same thing - and in yet another, they might be considered interchangeable at a certain ratio, or perhaps in a certain direction, or at a certain rate. It's all about what you're trying to do.
4. How is that equation useful? That's the only thing that matters about both definitions and equations. I can define a million things that are completely useless if I wanted to - but what's the point?
5. Within the original context, those aren't considered at all. Within a more realistic context, both heat and sound are also kinematic.

The reason you have so much trouble finding the answer to your questions on physics sites and forums is that the question doesn't make much sense in physics. It's more about the philosophy of science, and the idea of building models of the world that try to describe reality to an approximation that happens to be useful to us. You think that those words have an inherent meaning that is applicable in any possible context - this simply isn't true. From the very inception of the idea of physics, people have known that it isn't (and never will be) an accurate representation of reality; and we've known for a very long time that, for example, different observers may disagree on the energy of one object. You just need to understand where a given model is useful, and pick the right model for the job. Don't try to drive a screw with a garden rake.

• That is really an awesome answer. I do not accept it because as far as I know, when you get few good answers, simply upvote them all and don't accept any in particular, because they complete each other. So in short, can I say that $W=F_{x}\cdot x$ is true only for kinematics? – Sipo Aug 31 '16 at 13:32
• @Sipo Well, not only in kinematics, but yeah, do not assume that definitions are universal. Just like W doesn't necessarily mean work in all contexts, and F doesn't necessarily mean force (and even . doesn't necessarily mean multiplication). – Luaan Aug 31 '16 at 14:08
• @Sipo: When you get good answers, by all means upvote them, but be sure to accept one of them too. Throw dice if you cannot choose which one. Accepting an answer is important because it will mark the question as resolved-to-the-asker's-satisfaction in the various lists of questions on the site. This is more important than maintaining theoretic fairness in the distribution of imaginary internet points. – hmakholm left over Monica Sep 1 '16 at 18:13

If, a man [has] 100J of energy, [can] he apply a force of $10^{100}N$ over $10^{−98}m$ ?

Yes.

As Archimedes famously noted, he just needs a very long lever, a suitable fulcrum, a long enough amount of time, and a suitable track.

The important point is not that all possible ways of expressing a law are feasible, just that all those that actually occur are observed to obey the law.

• There are practical limits. No lever can be built that would stand up to $10^{100}$ N. And things change at small scales. You need quantum mechanics to describe what happens. $10^{-98}$ m is below the Planck scale, so it gets more complicated. – mmesser314 Aug 31 '16 at 11:33
• @mmesser314 Well, the whole question is within the context of newtonian physics and spherical cows, which is an approximation where RedGrittyBrick's answer is perfectly fine. – Luaan Aug 31 '16 at 12:03
• @sipo this is not a reference to an experiment archimedes did. It is a quote: "give me a fulcrum and I will move the earth." This was after many experiments observing how levers work. He realised that with an appropriate lever he could move any mass, at least in principle, and this was his clever way of summarising it. – Level River St Aug 31 '16 at 23:24
• in greek Δος μοι πα στω και ταν γαν κινασω . "Give me where to stand and I can move the earth" – anna v Sep 1 '16 at 18:19

[Does] the definition of energy [mean] that every [form of] energy is potential energy?

It turns out it is not convenient or very useful to think like this. If you do then the word "potential" ends up having no real utility.

In a comment, tfb linked Feynmann on this subject

There are many places Dennis can hide the red cubes

• in motion (kinetic energy).
• in a heavy weight balanced on top of a cupboard.
• in a compressed spring.
• by moving electrical charges between two adjacent conducting plates.

It is handy to have different names for different places that Dennis can hide red cubes.

If you say that all forms of energy have potential to do work, you may be right, but it is still useful to have some separate names for forms we can distinguish.

It is the force, not the Force, you can feel and measure but it refers only to one actual setup, which is useless when looking for general understanding. Work, pressure, energy,... All that refers to forces in wider context.

Work
Analysing formula $W=\int_s\vec F\cdot d\vec r$ we can read how was force applied consumed to move the body over the path $s$. Suppose we are sliding a box on a pavement. Suppose the force is described as $F=\mu\ G$, where $\mu$ is coefficient of friction, $G$ is gravitational force. In that example, twice as much work is done to move two boxes over the same path (Force is doubled). Or we can slide two boxes half the way, compared to one box, for the same work to be done.

Energy
Energy is the ability, or potential, to do a work. Potential energy is one form of energy.

• Kinetic energy is energy stored in the motion state of the body and can be released by decelerating the body. You can see the body moving = it has energy.
• Internal energy is energy stored in thermal motion of particles inside the body and is percieved by the temperature of the body. If the body is gas, the energy can be released by moving the piston. You can feel the body is hot = it has energy.
• Electrical energy, also known as electric potential, is energy of charged body in given electric field. It is simillar to mass in gravitational field. Simpest example is charged capacitor. If it is discharged through motor it will move some body and the energy will be consumed. You can measue nonzero voltage = there is energy between the electrodes.
• Chemical energy, also known as reactional enthalpy, is energy stored in the difference between energy states of reagents and products (suppose the reaction is exothermic). This energy is released as heat - power source of thermal engines and explosives - or as a voltage - accumulators. There is label "explosives" = there is energy inside.
• Potential energy is quite tricky named. Imagine there is a concrete box. It is not moving (no kinetic energy), it has same temperature as its surrounding (no internal energy to be extracted), it is not charged (no electrical energy) and it is chemically stable (no chemical energy). Still it can do a work, because it is 5 meters above the ground. This concrete box has potential to gain kinetic energy that is able to do a mechanical work on the ground.

Work, potential energy and kinetic energy were defined when other forms of energy were not discovered yet or the connection between them and the motion was not discovered.

1. Yes, energy in any form, is potential to do work. And energy of a system can be increased by doing a work on it.
2. Opeprations with energy have their limits too. You cannot discharge car battery drawing 1 TA of current, you cannot lift 10ton car by bare hands. If you want to exceed such limits you need to use "tools". If you wan to apply $10^{100}$ N somewhere you need gearbox with $10^{98}:1$ ratio. In that case you will apply desired force on $10^{-98}$ m track, but your hands will apply 100 N force over 1 meter track.
3. You can see the difference when stunt man standing on a roof (having potential energy) jumped into paper boxes (consumed that energy). You can see a difference between a car approaching a wall (having kinetic energy) and the result of the EuroNCAP test (energy consumed to plastic deformation). You can see pile of wood, emptied canyster of petrol and jackass with burning torch (having enthalpy to be released) and flying logs and the blastwave (enthalpy released as kinetic energy of logs, heat and soundwave)...
Changing mass to gain energy is possible in relativity, where energy is mass times speed of light squared. Man reduce his mass doing work by consuming glucose (solid) and producing carbon dioxide (gas) and water (liquid and gas), which he breath out or sweat.
4. As I wrote above, the work is product of force applied and the track where it was applied. On the contrary $E_k=1/2mv^2$ was derived as a work needed to accelerate the body from steady to the velocity $v$. $$E_k=W=\int_0^lF\ dx=\int_0^l\frac{dp}{dt}dx=\int_0^l\frac{d(mv)}{dt}dx=\int_0^vm\frac{dv\ dx}{dt}=m\int_0^vv\ dv$$ $$E_k=m\int_0^vv\ dv=m[1/2v^2]_0^v=1/2mv^2$$ Physic apply mathematical tools to describe real world, therefore every physical quantity has its unit - a measure. In other words if you read $l=5\ \mathrm m$, you read that length $l$ is five times longer than the SI standard of meter. Nothing more, nothing less. Joule was defined as newton times meter, period. Another "rule" is to describe the object as simple as possible. Your definition $W=2Fx$ is not simple. To be valid it expects using meters and newtons as units of length and force, respectively, but halfjoules as units of energy. Using your definition, the kinetic energy gained by this work will be $E_k=mv^2$. On the other hand, if you measure work in joules you cannot use the formula using feet and pound-force - you have to convert them all to imperial units, or SI units. Nothing in between is acceptable.

1. Work is force applied over a track. That implies the motion by definition. Any force was applied on thet track means work was done. Laws of conservation and Newton's third law state, that energy cannot disappear, it can change its nature only, and where work is done ($\int F\ dr>0$) same ammount of work is consumed ($\int-F\ dr>0$).
Heat is energy transferred from point A to point B, percieved by change of temperature and performed by change of potential and kinetic energy of particles. Kinetic energy refers to velocities of particles and potential energy refers to their bonding states. Sound waves are periodical changes in pressure in fluids or displacement in solids, again on atomic scale. In both cases some particles do work on the particles. In electromagnetic field charged particles do work on another charged particles. But it is useless to describe some phenomenon by decribing each and every particle involved. When statistics was applied, temperature, pressure, heat, enthalpy, entropy and many others were defined.

Why $W = F_x \cdot x$ and not $W = 2F_x \cdot x$ ?

The constant is determined by our choice of units, it is just there so that the numbers we calculate correspond to markings on our rulers.

For example if $W = F_x \cdot x$ when $x$ is measured in yards, it is simultaneously true that $W = 3F_x \cdot x$ when x is measured in feet.

Since our choice of measurement scale is arbitrary (the length of some persons arm, the diameter of a certain planet, ...) the number is not very interesting. So it is sometimes omitted when discussing concepts, principles, etc.

It is convenient if you can choose a system of units so that this constant is 1 and therefore goes out of sight. Sometimes people do this.

• Writing $W=\mathbf{F\cdot x}$ expects using units of same origin, for example $Nm$, $N$ and $m$, respectively. $W(lbf\cdot yd)=3F(lbf)\cdot x(ft)$ is valid, because you define all units strictly, otherwise one expect $lbf\cdot ft$ as unit of work and the equation is invalid. – Crowley Aug 31 '16 at 12:07

As long as one remains on the phenomenology of Newtonian mechanics, I trust that your definitions are not wrong( I am of the opinion, and that's strictly a personal philosophical view, that there isn't a finite amount of definitions we can give to something, that could completely define it in an absolute manner). As for your questions:

1. Firstly, I am not sure if by potential energy you mean a potential energy function, such the one we can define for some forces( those that have $\nabla \times \bar F = 0$) as $\bar F= - \nabla V(r)$, or if you just mean linguistically the potential possibility under some circumstances to produce work because the object has some energy. But, staying clearly on Newtonian phenomenology, I'd say that having energy means exactly that you can cause a change, and that change may be seen as work.
2. I don' t think speaking of a man storing energy does any good in a scope of physics considerations. Let's just say that an object has some energy( for example kinetic energy or gravitational potential energy for being on a high h from the ground were we define U=0). First, there is no problem of acting with such a tremending force because force is given as the time derivative of momentum, a measure of the rate of change of momentum, or simply by considering that the mass of the object were the force is being applied is constant, $\ bar F= m \bar a$, where a is the acceleration. So, if an object has very little mass and I have a very big mass with a kinetic energy of 100 joules, I could definitely put a huge force on it because of momentum conservation during, let's say an impact. But let me also note that even if you have a quantity of energy Q, then the amount you can spend as a ratio of time is given by power, the time derivative of work.
3. Mass equivalence with energy does not bother us here because we are in the field of Newtonian and not relativist mechanics. An object having an kinetic energy T is distinguished by an another with Q by their observed difference in momentum. Two object defined in the same frame of reference with some potential energy differ by the kinetic energy they could gain and from the impact they could thus have. Different potential energies mean different kinematics.
4. The definition is simpler by not having the two in front. Why should we need a factor that would just give as a different scale in measuring? Work is defined as scalar of two other vector entities, so why have a multiplier if it doesn't do us any convenience?
5. You could again define the force a body will get by it's momentum change, so that a mechanical wave having some energy means it could pass an amount of that energy to another object, that changing it's momentum and affecting it's kinematics. That is doing some work. Heat is something different since it is not consider as useful as other forms of energy. But, thermodynamically speaking, suppling a system with heat leads to an increase of it's temperature and thus to a statistical increasing of the kinetic energies of the system's parts. If the system had a fixed volume V and pressure P, and if we led it change it's volume without giving any heat to the environment, then the system, would increase in volume for a decrease in pressure, that giving an amount of mechanical work.

Hope this helps.

• in potential energy, I'm sure you meant "those that have $∇ \times \bar{F} = 0$" – WorldSEnder Aug 31 '16 at 12:59

Every theory of physics ends up being a description of a mathematical system. Just like any other such system (integers, real numbers, set theory, …) we can ask “why these definitions?” and the answer is necessarily “because we found this particular system to be useful for something” — for physics, predicting the behavior of physical objects.

In my opinion, the simplest such justification for the Newtonian concepts of energy and work is found by starting with energy:

Energy is a conserved quantity.

That means that we can look at any system at some moment in time, compute the total energy in it, and assuming we have not missed any of the forms of energy, then without predicting the behavior of the system in detail, at any future or past time, the total energy will be the same as it was now (plus or minus any in-flows or out-flows of energy, if the system is not “closed”).

This is sufficient to make energy interesting to physicists. It is also interesting to engineers and in everyday life because in order to do useful things you need to use some energy. (To understand why this is so, you will need to study thermodynamics.)

To summarize: Our definition of “energy” refers to something both theoretically meaningful and practically useful. That's all we really need “behind” it.

“Work” (in the formal sense) is not nearly as fundamental. It refers to just one of the ways energy can move from one part of a system to another. You could replace the word “work” with “mechanically transferred energy” and still be saying the same thing.

Does the definition of energy mean that all energy is potential energy?

Suppose you want to build a machine that throws a ball when a signal is given. Here are two ways you could do it:

• Compress a spring, hold it back with a latch, and put the ball in front of the spring. At the right time, release the latch and let the spring strike the ball.
• Spin up a flywheel. At the right time, drop the ball onto the edge of the flywheel.

Both of these devices put kinetic energy into the ball from some storage of energy. In the first case, the source is the “elastic potential energy” in a spring. In the second case, the source is the kinetic energy in the flywheel.

My point here is that even though it is often though of that way, “potential energy” does not actually mean “energy that is able to do work later”. Any kind of energy can be stored. The actual meaning of “potential energy” is “energy that can be calculated in terms of a potential” and the idea of “a potential” is not relevant to your question.

I defined energy as a property, or state of an object representing the amount of work it can do. But how could we quantify it?

As I said at the beginning, I consider energy to be the more fundamental concept, so this is not really a meaningful question given that premise. Instead, an amount of work is a change in energy — so it is quantified with units of energy.

How potential energy is reflected while being potential? Can we see a difference between a person that stores x energy and a person consuming that x energy?

You stated at the beginning that you wanted to stick to kinematics and Newton's laws. In that world, you can't just observe a quantity of energy; you have to total up all the forms in which it exists (gravitational, chemical, thermal...).

An answer I heard a lot is that the mass of the person is actually its potential energy, and therefore doing physical activities, for example, is consuming energy, and reducing the mass.

Mass–energy equivalence is a matter beyond Newtonian mechanics. You can't make use of it without stepping outside of the Newtonian framework. But let's do that for a second for the sake of answering your question:

Mass and energy (not just potential energy) are the same thing. The reason this isn't ordinarily obvious is that they are related by a very large constant, $c^2$. Therefore an everyday sort of change in energy causes an almost undetectable change in mass, and a noticeable change in mass corresponds to an enormous amount of energy.

But if it is correct, wouldn't we measure energy in kg or mass in Joule?

You could. The practical problems are that

• You'd have really small numbers for mass or really big numbers for energy.
• Even though they are the same thing, it is useful for practical calculations to maintain a distinction between mass and energy, and consider it a mistake to convert them without intending to by multiplying or dividing by $c^2$.

(Physicists often do work in alternate systems where mass and energy have the same dimensions; this general topic is called natural units.)

Although my definitions do explain the inner nature of work and energy, I could still mathematically describe this nature in an infinite number of ways. For example, if I describe work as $2F_x⋅x$, I would still get to the conclusion that work is the influence a force has on an object's displacement. So why is that the equation?

Because energy is conserved, we can do things that convert energy to other forms and conclude that there was the same amount of energy in the two forms (± any losses in the conversion mechanism). So if I define "work" as “$2F_x⋅x$” then I will either later find that I have to multiply “work” by 1/2 to relate it to something else (which suggests an obvious simplification), or end up with an unnecessary factor of 2 in front of all my formulas for calculating energy (at which point it makes more sense to use a unit of a different size so I don't have to write 2 all the time).

How is consuming energy in other forms than motion work? How are heat energy, or sound waves, for example, work?

In general, as I said at the beginning, I think it is not a good idea to see work as fundamental. But for those particular examples, there is identifiable work:

• Heat is (more or less) random motion of atoms. But “random” isn't fundamental — it just means that the motions are not practically predictable — not correlated with each other or anything else. Suppose you make some heat by rubbing your palm on a surface — you pushed against the atoms in the surface and moved them a little bit, doing work on them. The reason we call the result “heat” rather than “displacement” is that the individual displacements are all different and tiny.

• Sound waves are periodic motions of air (or anything else). Again, there is a displacement: it's just very small and repeated lots of times.

How [is] consuming energy in [...] forms [other than] motion [also] work?

When you do work on something, you change the amount of energy associated with the subject of the work, the energy is transferred from elsewhere.

so you can say that $W = \Delta E$

In the case of motion, the change is in the subject's kinetic energy.

In the case of a hoist, the change is in the subject's potential energy.

• But if work is $F_{x}\cdot x$, where is the $x$ for heat for example? – Sipo Aug 31 '16 at 10:29
• @Sipo, consider a fire-piston. $x$ is a displacement of the piston. The work done ends up as very hot cotton. – RedGrittyBrick Aug 31 '16 at 10:34
• But when the fire is already lightened up, the fire has heat energy there is no displacement. So how is there work? – Sipo Aug 31 '16 at 11:38
• @Sipo You're reading too much into it. Mechanical work is an idealized concept; even the notion of displacement is something that's very tricky to define precisely with what we know about how reality works. When you're working with a model, don't be surprised when the model starts giving the wrong answers when you step outside of the model. Both mechanics and thermodynamics have the same underlying reality that's neither "mechanical" nor "thermodynamical", but you're not going to design a lever based on quantum electrodynamics, are you? You just need to understand the limits of the model. – Luaan Aug 31 '16 at 12:07
• @Sipo The easiest way is probably to start from a good description of physics as a subject - any good textbook will do. Feynman's Caltech lectures (for energy specifically, feynmanlectures.caltech.edu/I_04.html#Ch4-S1 stands pretty well on its own) are a great introduction to physics. – Luaan Aug 31 '16 at 12:57