# The Physics Philes, lesson 13: The Crown Joules

*In which no work is done, kinetic energy is explained, and I am rolling in joules.*

So far in our little physics adventure we’ve been talking about the forces exerted on objects and particles. But now we’re going to switch gears a little and discuss work and kinetic energy. But what do I mean by work and kinetic energy? The definition of work is different from how we use the word colloquially. For example, we might say that, as you study for a chemistry exam, that you are working very hard to get a good grade. However, in physics terms, you probably aren’t doing very much work at all (unless studying includes hauling boxes of Erlenmeyer flasks and Bunsen burners from one room to another). Or an equally matched tug of war. If you may be working really hard to pull the other team into the mud, and you may get pretty tired. But as long as there is a stalemate, neither side is doing much work.

To do work in physics, a force must be exerted that causes a body to be displaced. The unit of work we’ll use is called the joule (J), and it’s the force exerted times the distance the body moves.

1 joule = (1 Newton)(1 meter)

When the work being done is the result of a constant force in the direction of a straight line displacement, the work done equals the force exerted times the distance:

W = Fs

Where W = work, F = force, and s = displacement. But what about when the force exerted is exerted at an angle with the body’s displacement? In that case, we have to dig up what we’ve learned about vectors. The force vector will have a component that is parallel to the displacement, and one that is perpendicular to the displacement. For the parallel force:

**F** = FcosΦ

For the perpendicular force:

**F** = FsinΦ

However, the force acting perpendicular to the direction of displacement isn’t doing any work on the body. That means the work done is equal to (FcosΦ)s, or:

W = FscosΦ

If the force component is in the direction of the displacement (between 0 and 90 degrees), then the cos is positive and the work is positive. If the component is in the opposite direction (between 90 and 180 degrees) then the work done is negative. If the work component is perpendicular to the displacement, then the work done is zero.

The concepts of doing positive and neutral work are pretty easy to grasp. Either the object moves or it doesn’t. But what does negative work mean?

Let’s think about lifting weights. When you lower a barbell, your hands and the barbell move in the same direction. But your hands also exert a force on the barbell that keeps it from falling to the floor (Newton’s third law of motion!). For force your hands are exerting on the barbell is negative work.

The word equations that we’ve looked at so far only do one force at a time. What if there are lots of forces acting on a body? You could calculate the work for every force individually and add them together. Or we can use what we know about vector addition to add the vectors together and using the vector sum to calculate the work. Either way works.

There are a few things we know about what happens when a particle experiences work. When a particle undergoes a displacement, the particle will speed up if the total work is greater than zero. The particle will slow down if the total work is less than zero, and the particle will will maintain it’s speed if the total work equals zero. Let’s try to explain this in math.

Let’s say a particle is moving at a constant speed along the x-axis. The constant acceleration can be expressed with Newton’s second law of motion: F = ma. Let’s further say that the speed changes from one point to another (call them v sub 1 and v sub 2) and undergoes a displacement of s = x sub 2 – x sub 1. With these values we can use one of the constant acceleration equations we learned earlier.

We can replace v sub x with v sub 2, v initial with v sub 1, and (x – x initial) with s. When we do, we get:

That last equation is the equation for acceleration. If we multiply that by mass, we can calculate the force:

The value 1/2(mv squared) is the value we give to kinetic energy. Kinetic energy is never negative. The particle either moves (in which case the value is positive) or it isn’t (in which case the value is zero). The work done by the net force on a particle equals the change in the particle’s kinetic energy:

Voila! The work-energy theorem! Like work, kinetic energy is also measured in joules. Because the work-energy theorem is derived from Newton’s second law of motion, it will only work in an inertial frame of reference, and the theorem was derived for the special case of straight line motion with constant forces. So, you know. There are limits.

All of seems to be pretty straight forward. I mean, even I understand the math, so how hard can it be? But what about complex systems that can be represented with particles with different motions? Well in that case, it gets more nuanced and I’m not going to get into it right now. (Don’t worry. I will eventually.) Just know that there are things you can do for that situation.

While we’re discussing things that are over my head, I’d like to talk to you about what happens when work is done, but the forces vary. I say I’d like to, because I can’t. Not now, anyway. Don’t look at me like that. I’m trying! Okay?

Now that that’s out of the way, we need to discuss one last thing. So far we’ve looked at work and kinetic energy that has no relation to time. If a 10 N particle moves 1 m, the work done is 10 J, regardless of if it took 10 minutes or 10 years to do. But when we talk about power, we add the element of time. Power is the rate at which work is done. The average power is amount of work done over a period of time:

To get the instantaneous power, we calculate at t approaches zero:

The unit for power is the watt, which is one joule per second.

Whew! Explaining work, kinetic energy, and power was a lot of work! (See what I did, there?) Next week we’ll find out how the equations I’ve tried to explain today are actually applied. In the meantime, did I miss anything or mess anything up? Leave any advice in the comments!

*Featured image credit: joshwept*

## 3 Comments

Well, you did leave potential energy out, a very important concept if you want to apply it to problems.

Patience. I’m going by the syllabus 🙂

I am also working under the assumption that the syllabus outlines the best way to learn this stuff. I could be wrong. And if there is one thing this project has taught me, it’s that I am wrong A LOT.