There are many questions on this site about the confusion caused by the definition of work done $W = \vec{F} \cdot \vec{s}$. Look over to the related column on the right you'd see how many there are:

I am glad the users are gracious enough not to mark them as duplicates, and I hope this hospitality can extend to my question which may result in opinionated answers.

Since most of the confusion surrounding work done is related to that, despite having a unit of energy, it does not provide much useful information regarding energy state of the parties involved. For example in the first link it was asked "How is the work done to push a planet over 1m with 1N the same as pushing a feather over 1m with 1N?" I might as well have asked about the work done of applying a force of 1mN to a particle of any mass travelling at 0.9c for 1s, because contrary to intuition implied by work done, displacement, absent from a force field, means nothing to energy state or energy transfer, and yet work done is defined with respect to displacement.

In the rare occasions that work done is used in a constructive manner, such as inside a gravitational field or between capacitor plates, it is being equated to other energies (KE, PE), whereas those energies on their own are sufficient to describe the energy states and the use of work done is redundant.

The question: In what scenarios is "work done" uniquely useful both as a concept and as a practical tool aside from the mathematical definition and in examinations? Note that I am seeking insightful examples as to when work done is a useful definition to assist problem solving (outside textbook exercises) without causing more confusion than problems it can solve. Like all definitions, there are useful ones, and not so useful ones. The vast number of work-related questions on this site alone proves the utility of this question. The philosophical aspects has been touched upon in this answer.

I am not disputing its validity. Please do not rehash the definition.

  • $\begingroup$ Surely it at least provides an easily calculable lower bound on the energy requirements to complete some process? $\endgroup$ Commented May 24, 2017 at 12:07
  • $\begingroup$ Work is just a mean of energy transfer - like heat. $\endgroup$
    – Steeven
    Commented May 24, 2017 at 13:01
  • $\begingroup$ @thomas-russell It isn't taught as a lower bound, is it? $\endgroup$
    – Moobie
    Commented May 24, 2017 at 13:38
  • $\begingroup$ @JMac I am looking for specific use cases, not a philosophical debate. I even said it in bold. $\endgroup$
    – Moobie
    Commented May 24, 2017 at 13:38
  • $\begingroup$ @MobileComputing You also linked to that question and it's answer as an answer to this question, so I assumed it was a duplicate as you personally seemed to think it was sufficient as an answer. $\endgroup$
    – JMac
    Commented May 24, 2017 at 14:04

1 Answer 1


If you agree that the definition of $E=K+P$ is useful and meaningful then you must agree that the work $W$ definition is at least as useful, since $\Delta E=W$. The definition of work $\vec F\cdot d\vec s$ brings always similar reactions at first because it doesn't seems particularly meaningful in itself, but it reveals his meaning in the relation above. Through this definition of work you can define $K=\frac 12mv^2$ and various potentials energy (notice that the definition of kinetic energy and of the common potentials are strictly related with the definition of work through $\Delta E=W$, defining the firsts directly brings to that definition of work and vice versa) and this quantity have the experimental and incredible property of being always conserved. So why we define work that way? Because $K=\frac 12mv^2$ happens to be a conserved quantity in our universe (Newtonianly speaking) and $W=\Delta K$. If we had experimentally noticed that defining $K$ like $mv^3$ would have been more physically meaningful and that was the quantity being conserved in interactions then probably we would have defined work in a different way.

  • $\begingroup$ I like your line "defining the firsts directly brings to that definition of work and vice versa" which brings about circular definitions in physics. But then it's fine since both quantities are describing the observed laws of nature. $\endgroup$
    – Moobie
    Commented May 24, 2017 at 13:07

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