Definition of Work — Physics
I believe one of the main reasons many people don’t get into pure sciences, especially math and physics is because we are not explained basics correct at lower schools. Today I will talk about the concept of work.
We are taught a rather circular definition of work. Work is energy transferred by force; and energy is capacity to do work. I am okay with this circular definition provided the equations that describe the same are explained correctly.
The first definition of work we learn is : work is force times distance. Note that we are given this as a definition without any fair justification into why that is true. By this blog post my intention is to give an explanation for that definition.
Force is defined as F = ma. I will not explain this equation in this post, but let us take if for granted. So work is W = FS = maS. Where S is the distance (or displacement).
It is easy to understand why work is proportional to mass. If we had 10 Kg to carry to a particular distance, and we split into two 5 Kg each. We would still end up doing the same amount of work. Similarly if we have to carry a 10 Kg twice to the same location it is same as carrying 20 Kg once. So we are very clear about the dependence of mass. Twice the mass, twice the work. Half the mass, half the work. So work is proportional to mass.
It is also easy to understand why work is proportional to distance. Carrying 5Kg twice is same as carrying 5Kg twice the distance. Similarly if we have to carry something twice the distance, we have to spend twice the amount of work. So again the relationship is clear, half the distance it is half the work, double the distance it is double the work.
But dependence on acceleration is not so easy to argue about. Because there is no reason to say why having twice the acceleration will cost as exactly twice as much? Why not 4 times? The argument is a little more subtle, since work is about how much “energy” we imparted on the box. But we don’t know equation of energy (yet).
But we know it should be intrinsically a property of the box we are pushing. (We are living in a “frictionless” world, aka friction is nothing but electromagnetic forces at microscopic scales, so everything is still consistent). The energy of a box cannot depend on where it is located, it will be same wherever it is. Energy cannot depend on acceleration, because it is then acquiring or losing energy. If nothing is acting on the box it will stay at uniform motion. So it is reasonable to expect energy is a function of velocity E = f(v) given everything else of the box is a constant (like mass, ideally it should increase mass when you push, but let us ignore that for a moment, we are in Newtonian world right now).
Since v² = u² + 2aS ; in order to get same end velocity at half the acceleration we need to double the distance. So in other words half the acceleration with double distance is same work as before. We already know halving the distance will halve the work done. So half the acceleration with same distance must be half the work done. Similarly twice the acceleration with same distance will be twice the work done. Now we know work is proportional to mass, acceleration and distance. Hence W = m a S.
One could ask is there any variable we forgot to consider? Turns out no because velocity doesn’t depend on anything else other than acceleration and displacement. And mass is constant for a given box. Only when you cut the box into two it becomes two of equal sizes, and hence the dependence on mass.
If you can read this blog in 5 minutes, it is possible to explain this concept to high school students in one month. Understanding work is so critical, it is better to spend the time and explain why the equation represents what it does.
