# How is the concept of “work done” useful? [duplicate]

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.

## marked as duplicate by ZeroTheHero, John Rennie newtonian-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 25 '17 at 5:43

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.