# Is this definition of work right?

My thermodynamics textbook defines work as follows.

Work is motion against an opposing force.

But this definition of work doesn't imply that work is done in accelerating a body does it? So is this definition of work wrong?

• Which book is that? – CuriousOne May 18 '16 at 19:49
• This is a really bad definition. – Omar Nagib May 18 '16 at 19:49
• @CuriousOne Atkin's Physical Chemistry – The Cryptic Cat May 18 '16 at 19:52
• @CuriousOne Sure it can. It's just that we usually leave out the change in KE of the system when we apply, say, the first law of thermodynamics to a system. However, most texts at least start out by including KE in their formulation of the first law. See, for example, Fundamentals of Engineering Thermodynamics by Moran et al, or Introduction to Chemical Engineering Thermodynamics by Smith and van Ness. – Chet Miller May 18 '16 at 20:11
• @CuriouOne: That was not my point at all. I did not say that either the mechanical energy balance or the thermal energy balance could be used to model anything in science. But, I do assert that the differential overall energy balance that is used to analyze Boeing 747 flight is derived directly on the basis of the open system (control volume) version of the first law of thermodynamics, which includes KE. And the differential mechanical energy balance equation that is also part of the 747 analysis is obtained by dotting the equation of motion with the velocity vector, and also includes KE. – Chet Miller May 18 '16 at 22:49

The definition of work:

$$W=\int \vec F \cdot\mathrm{d}\vec x$$

So, work requires a force and a displacement. That is all. Think of it like this:

• If you push hard on a wall, you might use much effort to apply large force - but nothing moves and no work of use is done.
• If you push against a balloon, you can make it move very far. But you didn't really do much work because there was no effort in it since the force needed was so small.

In this sense, work makes sense in the every-day logical use of the term: You wouldn't call it work, unless it did a proper change (displacement) and required some amount of effort (force).

My thermodynamics textbook defines work as follows.

"Work is motion against an opposing force."

Simply untrue. Work is not motion (and what does that even mean? Meters per second?) I guess they mean displacement; but that is still an incomplete definition. Also, work is nothing "against" a force. A force might do work and forces might counteract each other - it doesn't make sense to say that work goes against a force.

But this definition of work doesn't imply that work is done in accelerating a body does it?

And why should it? Work has got nothing to do with acceleration. Looking at the formula, it doesn't really matter how fast or how much the object accelerated while you pushed it. The only thing that matters is how hard you pushed (force) and how far (displacement).

An example is you pushing a car up a hill. At constant speed. No acceleration, but certainly a force exerted and a displacement happening.

• if $W=\int \vec F \;\mathrm{d}\vec x$ and $\vec F=m \vec a$, then wouldn't $W=\int m \vec a \;\mathrm{d}\vec x$? I'm having trouble thinking of a scenario where there is work being done but there is no acceleration. Can you give an example? – Armadillo May 18 '16 at 21:21
• An example: Me pushing my car up a hill at constant speed. – Steeven May 18 '16 at 21:36
• @jakemcgregor The expression $\vec F=m\vec a$ is incorrect - or at least not valid in general. There is no law that says this. Rather the expression is: $\sum \vec F=m\vec a$. The sum symbol $\sum$ might feel unnecessary, but it is of huge importance. The point is that any force might do work, and they can even do that while balancing each other out, resulting in no acceleration $\sum F=0$. – Steeven May 18 '16 at 21:36
• was your car initially moving at the same constant speed uphill prior to you pushing on it uphill? For simplicity, I would think it (the car) would initially have a velocity of zero, and for you to get it to some constant speed would require you to accelerate your car, no? – Armadillo May 18 '16 at 21:41
• @jakemcgregor and Steeven re (3): the work-energy theorem and Newton's second law are mathematically equivalent. So no, you can't have nonzero net total work and zero acceleration simultaneously. Re (1): the confusion may come from the fact $W=\int F·dx$ is a mere definition, it cannot be true or false; whereas the 2nd law is a principle, it can be experimentally tested. – L. Levrel May 19 '16 at 9:33

A more general definition is ( the integral ) of force times distance.

A static force, eg a body resting on a surface , does no work.

The same body dropped has work done on it by the force caused by Gravity, potential energy being gradually converted to kinetic energy for the duration of the fall.

• Your answer is right but misleading: make it clear that acceleration is not necessary for work to be done (which the OP seems to think). Give an example where there's no acceleration! – L. Levrel May 18 '16 at 19:59
• @L.Levrel if $W=\int \vec F \;\mathrm{d}\vec x$ and $\vec F=m \vec a$, then wouldn't $W=\int m \vec a \;\mathrm{d}\vec x$? By definition, force causes only a change in velocity (an acceleration); it does not maintain the velocity, and for there to be work, there must be acceleration? – Armadillo May 18 '16 at 21:10
• @jakemcgregor: see Steeven's comments, which I fully support. – L. Levrel May 19 '16 at 9:44

This may not be a definition of work. You need to put on more context. If the author was talking about how to get the work calculated, it is fair to make this statement.