https://youtu.be/LSj91an4cjY
PHYS 1101: Lecture Sixteen, Part Two
So let’s get started with that. Let’s build up our understanding of what this new equation is. Well, here it is written at the top. This is our goal. This is our new starting equation that captures this energy conservation constraint we are going to work with to solve our problems. We have to know and understand how to use each of these terms or what to plug in. What do each of these terms represent? Here is how we are going to go about understanding that.
The first bullet, I want to point out, is let’s go back to our fundamental equation that we worked with last lecture, which is all of the work that gets done is the change in kinetic energy, and let’s break it down to show you the nature of this work on the left. For objects that are on the surface of the Earth, often times, gravity is going to contribute to this network if there is any up-down motion, you’ll see, and the amount of that work only depends on the height change of this object. It ends up being then a very easy work contribution to calculate. If you just take a little closer look at the nature of it and see how to do it, then you realize you can pull out that contribution and that’s what we’re going to do.
So let’s write, for any object on the surface of the Earth, all of the work that gets done is a sum of the work done by each force on the object. Some of these may be zero, as we have learned, some can be positive, some can be negative. If it’s on the surface of the Earth, we know there’s going to be the force of gravity on the object, and so I can have potentially work that this force does.
Here’s what we’re going to do. We’re going to replace this with quantities that are very easy to calculate and in essence, we are going to see that contribution, we are going to effectively move it over to the right side of the equation. Why are we going to do this? Well as I mentioned a moment ago, the force of gravity is always on an object that’s on the surface of the Earth. It’s always pointed down and because of that, because the nature of that force never changes, the work that gets done ends up being very easy to represent.
This force due to gravity is called a conservative force. Conservative means, as you’ll see, it captures the notion that, if we go from one position to another, and the gravity does positive work in that process, if we go back from that position back to the initial position, no matter what path we take, gravity is going to end up doing the same amount of work, just the opposite sign. In other words, the work that gravity does as we go back and forth between two positions, it’s always the same amount. You’ll see that in a minute.
Let me start to emphasize that by doing a little sketch and a little thought exercise here for us. We’re focused on the force of gravity, so here I’ve got it drawn and during this problem, let’s say some object undergoes the displacement shown there off at an angle called S. This literally would be, for example, a side view, perhaps a box sliding down a slope. There’s a real height change, a real altitude change by the fact that S is not purely horizontal, but there is some vertical part to it.
Well, I taught you last lecture, that to calculate any work that any force does, I have to multiply by the size of the force, mg, cosine of the angle, times the distance. Where the angle here would be the arc between these two vectors if I put their tails together. So this combination of terms, I emphasize for you, gives you the component of W, this force, in the direction of motion, which is what I’m highlighting here for you. This here, that I have highlighted, is the component of the force due to gravity, in the direction of S, the displacement. That’s what’s highlighted in blue, because it’s only that part of the force that contributes to this motion in that direction.
Well, I want to show you that if I just think of grouping this terms slightly differently, that I can interpret this in another way. If I group S times the cosine of theta, what that gives me is what’s shown in green here. This S times the cosine of theta is the component of the displacement in the direction of the force of gravity.
So pause for a minute and think about your trigonometry. Don’t you agree that what I have drawn here, the right triangle, the hypotenuse is the displacement? The hypotenuse times the cosine of the angle gives me the adjacent side, so this is the component of S that’s in the direction of W. So these are two physically equivalent ways of interpreting this mathematics, of translating this mathematics.
I introduced it by having you think about always the part of the force in the direction of motion, but it’s physically just as meaningful to think about the work as also being a measure of the amount of the displacement or the amount of the motion that’s in the direction of the force. If you do this with respect to gravity, then you appreciate that what’s going to happen with a force that’s always constant and in the same direction, is that only the overall motion in that same direction is going to contribute or count in terms of the work that that force does.
Here’s what I mean. Let’s say that an object goes from an initial height in position A to a final height in position B, so this is a lower altitude and it looks like it’s offset to the right. Let’s say this object gets there by undergoing a sequence of displacements. Overall though, I end up with a height change that is, if I assign a value to this altitude initially, and a value to this altitude at the end, then the difference between these two would represent my height change. Just the altitude change.
From what I just pointed out above, every time I undergo a displacement at some angle, the component, the amount that counts in terms of the amount of work that gets done is only the vertical component of that displacement. So for this first part of my journey, I travel this distance, so the amount of work that happens for this part of the trip is mg times this component, that height. As I then turn and head straight down, I get, pick up more contribution to the work that gravity does. I then take a turn to the left and undergo this motion, going down at a slope, but again, gravity only is going to do mg times this distance amount of work. And so on.
This keeps happening as this person, this object, meanders down and moves from point A to point B. In the end when I add up the work contribution from all of these segments, it’s going to be mg times this overall distance which is the sum of just all of the vertical drops that occurred. The work that gravity does is mg then times this overall distance d.
Notice that it would be positive because the force is in the direction of the general motion. The force is down, the motion is down, gravity is doing positive work. It’s adding energy working to speed up the object. We don’t know if the end result is that the object speeds up because we’re only looking at one contribution to the total work. Just the work gravity does.
But that contribution definitely would be positive and it very important that you appreciate that the calculation of that, it only depended on the height change. I could have imagined a different, very complicated path, but for every little segment of that path, it would still only be the vertical drop that I would end up adding up. Okay. But here’s how I want to write this so it’s a little more convenient. It’s going to be convenient in a problem to literally call out what your altitude number is. Call the lowest altitude in your problem zero if you’re at, you start out at a height above that, let’s call that positive number. Plus 5 meters. That would be you altitude above your zero. So these have to represent numbers. Literally tell us our altitude.
We’re used to in physics, thinking about a final altitude minus an initial altitude, so in order to write this value d, so I end up with a nice positive number, if I use these altitude values for what I’ve drawn here, I have to write it as minus the final height minus the initial. So I can have my final minus initial, which I would write as my delta h, but in order to get the distance to come out positive, I have to take the negative of that.
Do you see that? Because, for the scenario I drew, the altitude starts out higher, larger. This is a larger number, so this is going to be negative. I need to take the negative of that in order to get to a positive number. Here’s what happens. The work that gravity does then, instead of having to think about this d value and worry about the sign, I can use this form where I can literally plug in the altitude numbers, and I’ll get the right value for the work.
For example, if I had started out downhill and worked my way up, this would end up being a positive number, and the negative then would ensure that gravity does negative work in that process. As something goes uphill, gravity is working to slow it down.
Okay, so in the red box here is my summary of the work that gravity will always do for an object. An altitude change, if that occurs, I calculate the work that gravity does using this equation. Here is highlighted again for you in the box.
Here’s how to use it. Call your lowest altitude height zero. That’s convenient because then you’ll only end up working with positive numbers as you plug in for these two variables. The initial altitude, if it starts out that the initial to the final altitude means that you go downhill, when you plug in those altitude numbers, you’ll see that the work gravity does is positive. This physically captures the fact that that force gravity is in the same general direction as s, downhill. Therefore, gravity is working to speed up the object.
If your initial to your final altitude values represent going uphill, meaning Hf is larger. This is the positive number and that’s zero. When you plug those numbers in, you’ll see that the calculation will leave you with everything on the right is negative, meaning the work is negative. The left side of the equation then is negative. This has the physical meaning that if you go uphill, force of gravity is opposite to the general direction of the motion, so gravity is working to slow the object down.
So, there I’ve got it highlighted in green for you. If you’re ever asked to calculate the work that gravity does, this is the direct simple equation that you can use to do that. Follow the recipe here that the values for these altitudes comes from the literal height of the object at the final position compared to the initial. You want to call your lowest altitude zero. That might be at the start, it might be at the finish. The other one then is just going to have a positive number showing that the object was at a higher altitude at the other instant.
Okay. Let me show you that and use some real numbers here to see how to use. Let’s say an object goes from location a to location b so my initial altitude zero represents my height here. The variable hf represents my altitude here. I’m going to call hf zero. I’m picking my lowest altitude and calling zero.
Now we’ve learned that it doesn’t matter what torturous path we take as long as we started here and we ended up here, during this whole process, the work that gravity does, only depends on the altitude change. If you’re concerned about these cases where I started downhill for a while and then went back up, one way to think about it is, during a segment where I had a downward trend, the amount of work gravity did then ends up being just the negative of the work when I did the same upward trend. All of those downs and ups cancel to the extent that whatever my overall altitude change is, that’s what I’m left with.
So, I’m going to use my equation. The work that gravity does I can calculate as minus mg times h final minus h initial. Let’s just plug in numbers. I was given the mass, 3 kilograms. I have to multiply by 9.8 and then I multiply by these altitude values. My final height is 0. My initial height is plus 2. So I end up with minus 3 kilograms times 9.8 and then 0 minus 2. That’s multiplying then by minus 2. The negatives here cancel. I end up with a positive work value that is 58.8 joules. This positive amount does confirm what we expect. The math worked out with the right sign, meaning that, as I go overall downhill from start to finish, gravity is in the general direction of that motion, the same downhill motion. Therefore the overall work gravity does is positive and my math left me with a positive value.
Okay. Here’s a quiz question for you. Go through this exercise. Calculate the work that gravity does, but now consider that the object goes from location b up to a using these same values. Okay. So here’s what we’re going to do. Perhaps you have already experienced it. When you write down that equation W equals minus mg times h final minus h initial, that it seems awkward. That there’s so many negative signs running around. I’ve got this one, and then the change in altitude. So here’s what we do. Because the work that gravity does is always given by just this change in height, in science we define what we call the potential energy of an object. Let’s calculate or let’s define the potential energy, and I’m going to use the symbol PE for that. The potential energy at any instant to just be mgh. So at my initial instant, my potential energy here would be mgh zero and at the end, at that instant, the final instant, my potential energy would then be mghf.
If I define these just as these positive quantities mgh at the two locations, then I can define the work that gravity does in terms of these values. Let me show you that it’s minus. Let me just multiply it out. mgh final minus mgh initial. I’ve just multiplied through this factor mg times hf and then mg times this term and I end up with this. So these are nothing more than my potential energy terms. This is potential energy final. This is potential energy initial. So I can rewrite that equation as being the work that gravity does is equal to minus the change in potential energy.
I show you these mathematical steps and what we called a mathematical derivation so you can appreciate; the logic of math is represented or is consistent with the logic of nature and the logic of physics. So when I make these substitutions and logically conclude that the work that gravity does can be mathematically written like this, it means physically, it’s equivalent to that too. I can always think of the work that gravity does as being equivalent to the negative of the change in potential. If I go downhill, I’ve lost potential energy. This will be negative. The negative and the negative will give me a physical meaning that if I lose potential energy, gravity has done positive work. If I gain potential energy, gravity has done negative work and this gaining potential energy will be represented by this term being positive, so gravity will have done negative work.
So, this is what we want to get comfortable with. We’re going to use these potential energy notions. Just shut your eyes for a minute and physically try to lock into what it means. My potential energy is mg times the height. The higher I am, the more potential energy I have. If I go from start to finish uphill, my potential energy goes up. If I go from start to finish downhill, my potential energy decreases.
Here’s how we want to write it out then. This was our starting equation as of last lecture. The idea is still the case, the net work has to represent the work done by all forces on the object. Let’s break it up as we have done before. That net work is a contribution of every possible work contribution from every force. Work of friction, tension, potentially the work of gravity, and this is then equal to ke final, minus ke, kinetic energy initial.
Okay, I’m going to now leverage what I’ve just done with learning that I can replace the work that gravity has done by minus the potential energy difference or change. And now all I’m going to do is make the substitution and then I’m going to move that over to the other side. I’m going to write that, what I’m still left with on the left, well first, let’s substitute it in. My net work is the work that friction does plus the net work of the tension force may do and then I’m going to put in minus potential energy minus potential energy initial. So I’m just writing out that substitution step.
Now all I’m going to do is just move this term over to the other side. By doing that then, perhaps it doesn’t seem simpler, but the more you use it, you realize that it is. I now am going to end up with an equation that has the work done by everything except gravity on the left, and I’ve moved the gravity contribution to the right and I’ve replaced it with this idea of potential energy. Do you see that where I simply do algebra here and I add this whole term to both sides of the equation to move it over? And here I have it written out for us with some labels. So you clearly can see what you’re left with then.
When you do apply this to a problem in general, I may have fewer or more terms, or perhaps no terms on the left, but everything on the left that started out as having to represent all the work that gets done, now has to represent all the work except the work done by gravity. All of the forces except gravity are called non-conservative forces. That’s why we end up labeling the left side just overall as saying Wnc for non-conservative.
Then, whatever work gets done by these forces other than gravity, that has to equate to, not just the change in kinetic energy, which captures the speed, the change in the speed of the object, but also I have to include this change in gravitational potential energy. This then becomes my new starting equation. On the left, now, I’m going to label it work nc for non-conservative. This is the work of every force except gravity. That’s because gravity I’ve got represented now over here in terms of gravitational potential energy.
On the right, I have the change in kinetic plus the change in potential and here are the substitutions that you may need to make or what these terms mathematically represent. The kinetic energy is always the product of one half mass times the speed squared. This is our speed comparison information in a problem. How the final and initial speeds compare. The potential energy terms, each one is mgh where h is the final and then the initial height. That’s position change.