Gravitational Potential Energy - Online Tutor, Practice Problems & Exam Prep

Gravitational Potential Energy

Hey, guys. So for this video, I want to take a closer look at gravitational potential energy. Remember, this is one of the main types of energy within potential energy, which itself is a type of energy. Let's check this out. Remember that the whole point or idea behind the work done by gravity is that objects fall, gravity acts on them, and if they start to fall faster, meaning their speed increases, then their kinetic energy is going to increase. We've seen this type of situation before, but let's take a look at this example here. So the idea is that we have a 5.1-kilogram box that's falling from an initial height of 10 meters and it falls to a height of 4 meters, which means that it has a displacement, some delta y which we can calculate as negative 6. If we want to calculate the work that's done by gravity, right, as this thing is falling, we're just going to use our equation: -mgΔy. This is -5.1×9.8×-6. What you're going to have is the negatives cancel, and you're going to get the work done by gravity is 300 joules. It makes sense that the work done by gravity is positive because it's helping the box speed up as it falls. Gravity does positive work if it aids or accompanies the motion. Right? So remember, this work-done really changes the kinetic energy of the box. Here, the initial velocity is 0, which means that the kinetic energy is equal to 0. But when it's down here, it has some final velocity and the kinetic energy is just going to be whatever the work that was done was. So first, it goes from 0 joules, and now it's going to have 300 joules of kinetic energy because gravity did 300 joules of work on this box. So the point of this video is that this energy has to come from somewhere. It can't just be created or destroyed. We know that. And the idea is that this kinetic energy has to be transferred from another type of energy called gravitational potential energy. We've talked a little about potential energies before. We say that they're essentially stored energies, and that's exactly what this potential energy is. Gravitational potential energy, which we write as ug, is stored energy primarily due to an object's height. So the fact that this object is at a height of 10 meters means it has potential energy, some stored energy that can then become kinetic energy as gravity acts on it. Alright. So the formula that we use for ug is just going to be mgy. Some textbooks might also write this as mgh. So if you see that, it means the same thing as mgy. So let's go ahead and get to part a here. In part a, we're going to calculate the initial gravitational potential energy. So that's really just going to be u_g,initial. So if ug is mgy, then we're just going to calculate this by doing mgy_initial. We have all the numbers for this, 5.1, 9.8, and then the initial height is 10. If you go ahead and work this out, you're going to get 500 joules. So basically, because this box is 10 meters above the surface, it has some initial stored energy of 500 joules. So let's take a look at part b. In part b, now we want to calculate the final gravitational potential, which we're just going to use mgy_final. So this is going to be 5.1×9.8×4, and this is going to be 200 joules. So this makes sense because now when you're at 4 meters, you have less height above the ground, so you should have less stored energy. Your u_g,final is equal to 200 joules because you've lost some height. And now in part c, we want to calculate the change in gravitational potential. Remember, change just means final minus initial, so Δu_g. So Δ is just going to be final minus initial, so there's u_g,final minus u_g,initial. And we just calculated those things. Remember, u_g,final is 200. U_g,initial is 500. So basically, we have 200 minus 500, and this equals a change of negative 300 joules. Now this should make some sense that it's negative because as you're falling and losing height, you are losing some of that stored energy. So your Δu should actually be negative 300 joules here. But there's actually another way to understand what's going on here. So what I want to do is actually want to rewrite this u_g,final and u_g,initial, and basically replace them with their equations. This is mgy_final minus mgy_initial. So I can group together the mgs, and I can say that this is mg(y_final-y_initial). Now remember that final minus initial is just Δ. So really, this Δu_g is equal to mgΔy. So if you take a look at the equations here, we can see that this change in gravitational potential energy, which is mgΔy is the negative of the work that is done by gravity which is -mgΔy. These two equations are opposites of each other. And so basically, what we can say here is that the work that is done by gravity is the change, specifically, the negative change in gravitational potential energy. So w_g is -Δu_g. So really what we can do here is come up with a relationship between the work that is done by gravity with the change in the kinetic energy and the change in the potential energy. And, basically, what we can say is that if there are no other forces acting on an object, then as objects fall, then what we've seen is that the work that is done by gravity is going to be positive because gravity pulls things down, it makes them go faster, and it's going along with the motion. Your kinetic energy is going to increase because your speed is going to increase. But as you fall, you're losing height, which means that your potential energy is going to decrease. So your Δy is going down. When you're rising, when objects start to rise, then it's the opposite. The work that's done by gravity is going to be negative. Your kinetic energy is going to decrease because your speed is going to decrease as you're going up. But because you're going higher, your potential energy is going to increase. So your Δy is going up. So really what happens here is that when you have objects that are falling, the kinetic energy really gets transferred to gravitational potential energy and vice versa. And the thing that does that transfer is really the work that is done by gravity. So if the force of gravity does work and it's really just a transfer between gravitational potential and kinetic energies. Alright. So that's it for this one, guys. Let me know if you have any questions.

Gravitational Potential Energy is "Relative"

Hey, guys. Let's work out this problem together. So here we have a 2 kilogram ball that's initially 6 meters above the ground. So let's go ahead and take a look at part a here. We want to calculate the change in gravitational potential energy, so that's Δug. We're going to be using this equation right here for 2 different cases. What happens here in part a is that we have this ball that is 6 meters above the ground. So I know that this sort of level right here, this y value is equal to 6. That's my initial. The ball then is going to fall down to another height. It's going to fall down to a height of 3 meters. So here I have y equals my yfinal equals 3. What I want to do is I want to calculate the change in gravitational potential energy. So you've lost some height and therefore you've lost some gravitational potential energy. So how do we calculate this? Well, what happens is we're just going to use mg×Δy. But what happens in energy problems is gravitational potential energy is always calculated relative to an arbitrary reference point. What happens in part a is that we're choosing the ground, which is y equals 0, to be where the gravitational potential energy equals 0. So what we're doing here is we're saying if this is 6 and this is 3, the ground level is y equals 0 and this is where ug is equal to 0. Notice if you plug in 0 into this equation here, you're just going to get 0. So what happens is we can actually just choose our ground level to be wherever we want. So what happens? We're going to calculate the Δu which is going to be mg×Δy, but I'm actually going to write it out in the longer way. So I'm actually going to write it out as mgyfinal-yinitial. That's what Δy means. So what happens? We're going to get a mass of 2. We're going to have a g of 9.8 and then my final minus initial is going to be 3 minus 6. I'm ending up lower than I started so I should get a negative number here. So I get 3 minus 6. When you plug this all in you're going to get negative 58.8 joules. This makes sense that you get a negative number because as you're falling, you should be losing gravitational potential energy. Alright. So let's take a look at part b now.

In part b, we want to calculate the same variable. It's the change in the gravitational potential, but now we're going to choose our reference points, this arbitrary reference point to be somewhere else. So now what happens is we have the floor like this, we have the ball that's still 6 meters above the ground, but now what we're doing is we're sort of choosing this height here to actually be 0. This is where my y equals 0, and therefore, this is where my gravitational potential energy is going to equal 0. The ball is still going to fall 3 meters and so it's still going to end up at some height, but now we want to calculate the gravitational potential energy. Right? So what happens here, it falls to a height of y equals negative 3. So it's going to fall 3 meters, the Δy, the change is the same no matter how you set the numbers. So the change is still 3. And so we want to calculate the Δu. So now what happens is my Δu is going to be mg and then yfinal-yinitial. So what happens? We're just going to get 2 times 9.8 times negative 3, because what happens is we're going to get negative 3 minus 0. So that's our yinitial. So what happens here when you calculate this is you're going to get negative 58.8 joules again. So it turns out that in energy problems, whenever you're calculating the change in the gravitational potential, only the change in the heights is important. So you can choose your arbitrary reference point, your relative, you know, where y equals 0 to be wherever you want in the problem. That's actually not going to change what happens to your change in gravitational potential because it doesn't depend on the initial or the final height. What only matters is the difference between these two points right here. So the Δy was equal to negative 3 in both of the cases here, and so that's why we end up with the same negative of gravitational potential. So usually, one good rule of thumb is that if you know Δy, if you actually know the change in the height, you can set the ground level, right, where we set, you know, our ground level or our arbitrary reference point to be wherever you want, and that's where your ug is going to be equal to 0. Usually, what you want to do is you want to pick the lowest point of the problem because it's going to make your calculations a lot simpler. Alright. So that's it for this one guys. Let me know if you have any questions.