# What exactly is $F$ in $W = \int_{a}^{b} F dx$?

I am trying to teach myself some basic physics, here is something I do not really understand about the definition of work:

When moving from $a$ to $b$ (in one dimension), the work done is defined to be

$$W = \int_{a}^{b} F dx$$

Now, say, I lift something up from the floor to a table covering a distance $h$. The work done is supposed to be $mgh$. This result is obtained by setting $F = mg$ in the equation above, which is the force exerted by gravity on the object. However, gravity is not the only force acting upon the object. What about the force used by me to lift the object, why does it not enter the equation?

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The definition of work depends on which force is considered. –  Qmechanic Apr 25 '11 at 15:28

What I've been telling the students in the intro mechanics class I TA is that each force corresponds to some amount of work. The individual contributions of work add up to the net (or total) work, just as the individual forces add up to the net force.

So, for instance, when you lift a box up from the floor to a table, there are two forces acting on that box: the force $\vec{F}_\text{lift}$ you apply to lift it, and the force of gravity, $\vec{F}_g = -mg\hat{y}$. You can calculate the amount of work done by the lifting force as

$$W_\text{lift} = \int_0^h \vec{F}_\text{lift}\cdot\mathrm{d}\vec{x}$$

and you can calculate the amount of work done by the gravitational force as

$$W_g = \int_0^h \vec{F}_g\cdot\mathrm{d}\vec{x}$$

In a typical physics-problem scenario, you would be lifting the box at constant velocity (except for the brief moments when you start and stop its motion, which I will ignore for now). In this case, you could draw a free-body diagram and use Newton's second law $\vec{F} = m\vec{a}$ to discover that your lifting force has to be equal in magnitude and opposite in direction to the gravitational force:

$$\vec{F}_\text{lift} = mg\hat{y}$$

Armed with this result, you can calculate the work done by each individual force (try it!): $W_\text{lift} = mgh$ and $W_g = -mgh$. The net work done is zero.

You can just as well calculate the net work from the net force,

$$W_\text{net} = \int_0^h \vec{F}_\text{net}\cdot\mathrm{d}\vec{x}$$

In this case, the net force is $\vec{F}_\text{lift} + \vec{F}_g$, which is zero (because the box is moving at constant velocity, remember $\vec{F}_\text{net} = m\vec{a}$). So, again, the net work is zero.

"But wait," you say, "I thought the work done was $mgh$!" Well, it is, if you only consider the work you did to lift the box. This is what you would measure if you used a crane to lift the box, for example: the amount of work that needs to be done by the crane is $mgh$. However, gravity did the opposite amount of work, which canceled out the work you did (except for those brief moments at the beginning and the end, but those two contributions cancel each other out anyway). So the total work done by all forces involved is zero.

This might seem a little confusing - perhaps you're wondering, if the total work done is zero, how did the gravitational potential energy increase from $0$ to $mgh$? The trick there is that the "work done by gravity" $W_g$ is actually the gravitational potential energy under a different name. For conservative forces, we define $\Delta U_i = -W_i$: the change in a certain kind of potential energy is the negative of the work done by the force corresponding to that potential energy.

In this case, the gravitational force does work $-mgh$, so the change in gravitational potential energy should be $mgh$. But once you start calling it a change in gravitational potential energy, you can no longer call it a contribution to the work - and that leaves you with just the $W_\text{lift} = mgh$ contribution. So in that case, if you label the "action" of gravity as a change in potential energy rather than a work, the work done is $mgh$.

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You say "except for the brief moments when you start and stop its motion, which I will ignore for now", but this is a significant source of confusion for me; if I take an object at rest and lift it, the "textbook solution" says that I do enough work to overcome gravity for that distance ($mgh$), but that force alone applied doesn't guarantee motion, just constant velocity. Can you elaborate on this point that you elided? –  BlueBomber Nov 10 '13 at 8:50
@BlueBomber I mention later on in the answer that the work done cancels out between those two brief moments. Basically there are three "phases": (1) start the object moving at speed $v$, which takes work $\frac{1}{2}mv^2 + mg\Delta y_1$, (2) constant speed motion, which takes work $mg\Delta y_2$, and stop it, which takes work $-\frac{1}{2}mv^2 + mg\Delta y_3$. Given that $\Delta y_1 + \Delta y_2 + \Delta y_3 = h$, the total is $mgh$. –  David Z Nov 10 '13 at 9:15

The work by your lifting is equal and opposite the gravitational mgh work via Newton's Third Law.

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This answer is the only one to deal with the problem of the question: short and concise. Some of the other answers lack evidence for having understood the problem, maybe because it is too simple. –  Georg Apr 25 '11 at 18:38

I'm going to bet that you're being asked what the work done by nonconservative forces is.

Properly, you can imagine a conservative force being a force that satisfies $\nabla \times F =0$. If this is true, then the net work done by this force around any closed curve is zero. It turns out that there is also a central identity of vector calculus that tells you that if $\vec F$ is conservative, then for some scalar field $V(x)$, you have $\vec F=-\vec \nabla V$. Then, you start with the work energy theorem, and split your forces into conservative and nonconservative forces:

\begin{align*}\Delta KE &= \int \vec F \cdot d\vec x\\ &=\int (\vec F_{con} + \vec F_{noncon})\cdot d\vec x\\ &= -\int (\vec \nabla V)\cdot d\vec x + W_{noncon} \\ \Delta KE + V_{f} - V_{i} &= W_{noncon} \\ \Delta KE + \Delta PE &= W_{noncon} \end{align*}

Where we defined $\Delta PE$ as the total change in $V(x)$ over our path, and we used the fundamental theorem of calculus to do the integral defining the work done by conservative forces.

Therefore, the change in mechanical energy of the system is given by the work done by nonconservative forces. Since your system gains $mgh$ potential energy, it was necessary that external forces did $mgh$ of work.

But you are right in saying that the net work on the system is zero--the change in kinetic energy of the system is zero, so the work-energy theorem says that the net work done must also be zero. The work done by the gravitational force cancels out the work done in lifting the book. It's just that we rarely treat problems this way, preferring to account for conservative forces using the concept of potential energy.

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The force that you use is not included because the work is defined with the idea in mind to account for the work that you must put into a system to get it from $a$ to $b$. In the case of the table of height $h$, it is $mgh$, as you correctly point out.

Since you do not work against yourself, you don't include your own force.

The other point is that work done on a system is (at least in so-called conservative potentials) not lost, but stored in the object: If you lift your object to the table, you have done work $mgh$ and this is stored as potential energy in the object. If you let the object drop from the table, this energy gets converted into kinetic energy. At the bottom, you'll have $\frac{1}{2}mv^2 = mgh$, so you can calculate the velocity with which the object will hit the ground. The force that you used to lift the object does not contribute to this potential energy.

If, in addition to gravity, a friend of yours would also pull down on the object with a constant force, then you'd have to add the force he uses to the gravitational force in order to figure out your work.

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