A Bit on Work: Part I of III

The concept of work has a specific meaning in the sciences. It is best described by stating how it is calculated, which is the force applied to an object in the direction of its motion times the distance it moves. A force is, more or less, a push or pull. Newton stated that force, mass, and acceleration are linked through the equation F = m a. (He stated it in different terms, but this is how it is normally written today.) Some prefer to write it as a = F/m, to emphasize that a force causes acceleration and not the other way around.

If I push on something (say, a piece of paper taped to a wall) and it does not move, am I applying a force to the paper? Yes, but in this case I am not the only thing applying a force. The wall is pushing on the paper just as hard as I am but in the opposite direction. The net force F on the paper is zero and both the left and right sides of the equation a = F/m are zero and all is well.

Am I doing work on the paper? I'm applying a force but there is no displacement of the paper, so the answer is no. If I manage to push the paper through the wall then indeed I have done work (and I will feel very stupid). New example. Let's say a heavy box is at rest on an incline. I push on it with my arms parallel to the ground (not the incline), and it moves a bit up the incline. Let's say I push with a force of 20 newtons. (Newtons are the SI units of force. One newton is equal to the amount of force required to give a mass of one kilogram an acceleration of one meter per second squared.) Let's also say that the box moves up the incline a meter. Is the work done equal to 20 newtons x 1 meter? No, it's not, because we are concerned not with the overall force but with the force in the direction of the object's motion. We therefore need to mathematically subdivide this overall force (exerted by myself) into a component in the direction of the object's motion (which is of interest to us) and one in the direction perpendicular to this motion (which is not of interest to us). We then multiply the force component in the direction of the object's motion (which is going to be the overall force times cosine of the angle of the incline, with respect to the ground) with 1 meter to get the amount of work done. Yes, you can also calculate work by taking the total force I exert and multiplying this by the horizontal displacement of the object, which will be less than 1 meter. By the way, the SI unit of work is the joule, which is equal to, of course, a newton times a meter.

Let's see if I can confuse you. You push a bag of flour across the kitchen counter. According to Newton's third law, the bag of flour pushes back on you with an equal (in magnitude) but oppositely directed force. Why does the bag of flour move if the forces cancel out?

If you need a hint, take another look at the equation a = F/m. The forces exerted by you and by the bag may be equal in magnitude but your masses surely are not. Therefore you should have different accelerations, which indeed you do. This problem is complicated by the force of friction, which is robbing you of your acceleration and reducing the acceleration of the bag of flour. If you could get rid of the friction, you too would accelerate backwards (but not very quickly because of your relatively large mass). Do you see the difference between this example and the paper against the wall?

I want to talk more about work and a special kind of force that, oddly enough, never does any work. There aren't many forces in nature, so perhaps you can figure out which one I'm talking about. I'll discuss it in my next blog entry. Thanks for reading.