Work

"Work!" - - - Maynard G. Krebs

One of the words that appears both in everyday use and in physics talk is work; this is one that probably migrated from the everyday world into physics.

When we change the energy of a system by mechanical means, by pushing or pulling on it, we have done work on that system. The work we have done is just the amount of energy we have added. Yes, this means that if we've decreased the energy of the system, then we have done negative work on it.

Let's get to a useful equation for work. Start by thinking about a crate of mass m initially at rest on a frictionless floor. Starting at t=0 we apply a constant force F to the crate. We know that this means that the crate accelerates at a constant rate a=F/m. We also know that its speed, the magnitude of its velocity, is given by v=at=Ft/m, and that it has travelled a total distance x, which is one-half a t-squared. (Equations are kind of tough to insert.) Obviously, we can substitute F/m for the a in that equation. After a time t, we have done work equal to the crate's kinetic energy, one-half m v-squared. But we know v (it's Ft/m), so we can plug that into the equation for kinetic energy, and we find that W=(Ft)-squared/2m.

After twice as much time has passed, we've done not twice as much work, but four times as much work. Obviously, then, work is not proportional to force nor to time. We usually like to find that one physical quantity is proportional to some other quantity, or two, or three. But a little contemplation of the equations we've written so far leads us to notice that the work done in this case is equal to Fx. That is, the crate's energy increases by an amount equal to the force we apply to it times the distance through which it travels while we apply that force.

Notice that both things have to be there, both the applied force and the object's motion through some distance. If we stop pushing on the crate, it continues to move at a constant speed (after all, F=ma and now F=0, so a=0). Its energy does not change, and we are doing no work. Maybe more surprisingly, if we push on one side of the crate while the other side is in contact with a wall, the crate goes nowhere and we do no work. Sure, we may get pretty tired, but we won't have transferred any energy to the crate. (OK, not by mechanical means. We may have transferred some thermal energy to the crate, but that's not work.) Notice also that if we had pushed on the crate with a force that was not parallel to that (frictionless) floor, it would have accelerated at a rate lower than F/m, it would have gained less energy, and we would have done less work. Some profound thought should lead you to conclude that what matters is the component of force along the direction of motion. The vertical component of that force would simply have caused the floor to exert a greater normal force, and these two vertical forces would cancel each other and contribute nothing to the crate's energy.

You have to consider both the kinetic and the potential energy of the system. Suppose we steadily lift the crate through a height h. To do this, we apply an upward force mg, through a distance h, to the crate. The crate's potential energy increases by an amount mgh, and that is precisely equal to force times distance. If we had applied a force greater than mg, the crate would have accelerated as it rose, and thus gain kinetic energy, too; this makes sense because by applying a greater force, we'd do a greater amount of work on the crate.

If you stand around all day holding this heavy crate, you have done no work during that time. If you put the crate down at the end of this pointless day, you do negative work on it. If we've stopped pushing the crate across that frictionless floor, and it reaches a carpeted portion of the room, it eventually comes to a halt. Yes, negative work has been done on it, by friction.

What about forces that aren't constant, such as those exerted by a spring? When we compress a spring, we apply a force through some distance, but that force increases as the distance increases. Yikes! Is it OK to use the average force, and multiply that times the total distance? That's the sort of question that's easy to answer using calculus, and not so easy to answer otherwise. That's not surprising, given that Newton invented calculus just for doing physics. As we compress a spring by an amount x, the force we must apply increases from 0 to kx, where k is the spring's spring constant. The average force, then, is kx/2, and it's applied over a distance x, and  we'd estimate the work we've done by average force times distance, and find W=kx-squared/2. As it happens, this is the correct formula for the energy stored in a spring.