# Is work done defined for a state? [duplicate]

The following excerpt is from Energy principles and variational methods in applied mechanics, 2nd edition, by J. N. Reddy:

# 4.1 CONCEPTS OF WORK AND ENERGY

Consider a material particle, moving from point $A$ to point $B$ along some path in space under the influence of a force $\mathbf{F}$, which can be time-dependent. The position of the particle is measured from a fixed origin by position vector $\mathbf{r}$. Then the work $dW$ performed by the force $\mathbf{F}$ in moving the particle by an infinitesimal distance (or displacement) $d\mathbf{r}= d\mathbf{u}$ along the path over an interval of time $dt$ is defined as $$dW= \mathbf{F} . d \mathbf{u}=F_1du_1+F_2du_2+F_3du_3$$ In other words, work done is the product of the displacement and force in the direction of the displacement. The total work done, $W$, by the force $\mathbf{F}$ in moving the particle from point $A$ to point $B$ is given by $$W=\int_A^B \mathbf{F} . d \mathbf{u}$$ By definition, work done is a scalar quantity, and it is positive whenever both displacement and force have the same direction and negative if they hare in the opposite directions. Since $\mathbf{u}$ depends on the chosen reference frame, $W$ also depends on the choice of the reference frame. Thus, work is a relative quantity. However, work done does not depend on the path but only the end points, $W=W_B-W_A$. If the reference frame is chosen such that $W_A=0$, then $W=W_B$.

What is he trying to say? Is he saying that work in general is a state variable? And has exact differential? I got confused! So far, all of books I have read, also all of my professors, and many other sources and expert people claim that work in general is path dependent and only in some special cases can be determined by means of initial and final states. Is there something here that I don’t consider? Is there any meaning for $W_B$? Even for those special cases we have for example:

$W=U_B-U_A$ (not $W=W_B-W_A$)

where $U$ is some kind of energy.

## marked as duplicate by sammy gerbil, ZeroTheHero, Jon Custer, Yashas, user259412Mar 29 '17 at 6:07

• Depends on the kind of force. For forces like gravity it does not depend on path but for forces like friction (kinetic) it does. – Suhrid Mulay Mar 28 '17 at 9:13
• Please type out material you want to quote instead of adding a screenshot, since a screenshot cannot be indexed by search engines. – ACuriousMind Mar 28 '17 at 10:23
• @ACuriousMind You are right. I will do so. – lucas Mar 28 '17 at 11:31
• @sammygerbil This is not a duplicate. The current question isn't about a special case. This is about a general argument. – lucas Mar 29 '17 at 2:40
• @lucas Answers to that question address the general case. This question does not seem to be asking anything new. – sammy gerbil Mar 29 '17 at 3:06

There are two kinds of forces:

1. Conservative Forces
2. Non Conservative Forces

Conservative forces are the forces whose line integral over any closed loop is zero.

Non conservative forces are the forces whose like integral over any closed loop is not zero.

If the force under consideration is conservative, then the work done is path independent.

If the force under consideration is non conservative, then the work done is path dependent.

Your mention of "state function" implies you are coming at this from a thermodynamic background. Indeed, in thermodynamics, work is not a state function because a "state" in thermodynamics is not a position in space and the "work" is not done against a conservative force field (or any fixed force field, really). What forces are involved in the work being done depends on the thermodynamic process happening, and is rarely known or indeed useful to know.

However, this is not thermodynamics, you're reading a text about classical mechanics, where a "state" is merely a collection of positions and momenta of some objects, mostly point particles, and where the work is being done by moving the particle from one position to another against (or with) a force field. If the force is conservative, then the work done will be independent from the path taken, and indeed work with respect to that force will be a "state function" on the states of classical mechanics, but this terminology is not commonly used. The necessary and sufficient condition for a force being conservative is that there is some potential scalar function $U$ such that $F=-\nabla U$, where the negative sign is pure convention.

• Thanks! I am aware of conservative forces. I am saying that work is path dependent itself and doesn't have exact differential. And so, we cannot have something like "$W_B-W_A$". For conservative forces, the amount of work can be calculated by means of initial and final states, but this fact doesn't change the path dependent intrinsic of work. I think work is always path dependent but in some cases the amount of work can be calculated regardless of the path. – lucas Mar 28 '17 at 11:29

That depends.

If work is done by conservative force then it is path independent. E.g, gravitational force,

If work is done by non-conservative force then it is path dependent. E.g, friction.