https://youtu.be/SZWc86U2p0o
PHYS 1101: Lecture Sixteen, Part Six
In this case, I got a particle that starts at point A in the drawing, and that’s at a height of 2.9 meters. It’s projected down this curved runway. When it leaves the runway at point B, the particle is headed straight up, so it continues straight up, but now the force of gravity slows it down, brings it to a stop. We can ignore friction and air resistance, and we need to know and find the speed of the particle at point A that leads to a final height here of 3.9 meters. Okay.
Again, energy conservation is the way to go. I recognize that because I have height, that’s position information and speed information. I know the speed here is going to be 0 when it reaches that highest point. I’m asked to find this initial speed.
Okay. Some of you might be concerned, wondering if we need to break the problem up because the nature of the motion seems to significantly change at this point, when the object loses contact with that track that it’s on. The answer is that you don’t have to break it up there. Here’s why.
Our energy equation that we’re using is valid, as long as we represent on the right, the specific potential energies and kinetic energies at these two instances. So, the right side doesn’t care about the details in between. The left side does care about the details in between, but what’s required? I only have to represent all of the energy that’s added to or taken away from this object between the initial and final snapshots. As long as you can calculate that and represent it, you can have the motion or the forces potentially change during the middle of these two instances.
For this problem, what are the forces in these two scenarios? Well, when the object is sitting on the slope, I know I’m going to have a normal force, but it’s not going to do any work. I have the gravitational force on the object, which does do work, but it’s represented here by the potential energy. When the ball leaves contact with the surface, then the only force on it is that due to gravity. But again, its work contribution is represented by the potential energy terms. In fact, from start to finish, the only thing I care about is the altitude change for that object, when it comes to the gravitational contribution.
So, I can, even though the normal force doesn’t exist beyond this point, I don’t care. It doesn’t impact this equation anywhere. So, I don’t need to worry about breaking the problem up. Okay. So, let’s forge ahead. The object we need to focus on is the ball. For the left side, I need to consider any work that’s done by forces other than gravity. I only have walked through the forces on my object here in my picture, and the only force doing work is gravity. So, there’s nothing left. The left side is 0.
Okay. I’ve rearranged it a bit and moved our next quiz question up into your field of view. We’re thinking about the left side of the equation. We’ve concluded that w, work due to non-conservative forces is 0. So, question 19 is: Is mechanical energy conserved in this problem? Okay. We’re ready now to think about the right side. Those would be… I’m going to have to plug in the kinetic energy values at the beginning and the end, the potential energy values at the beginning and the end. So, I know to do that. I’m going to need to think about the initial speed, the initial height, the final speed, and the final height.
Going to define our lowest altitude as being 0. The initial speed, that’s the variable, in fact, that I want. That’s what I need to find. What’s my initial height? The initial height is 2.9 meters. My final velocity, or sorry, final speed is 0. What’s my final height? It’s 3.9 meters.
Okay. We could work through this problem and continue on with these numbers, and we would get the right answer. We also… In fact, try that. Try working through the problem as I’m about to with these original numbers.
Let me point out that in a problem like this, I could also have started with the lowest altitude as being here, where all I’m worried about is the beginning height and the final height. I don’t care that the ball actually goes below that before going above that. Let’s call this 0.
Let me write this a little bit better. Let’s call this 0. If this altitude is our 0 height, then our initial height becomes 0. What would our final height be? If I place my 0 here, I need to have a value for this height that’s the altitude above that 0. Well, that’s going to be 3.9 minus 2.9 because what they give me in the problem are the heights compared to the ground, but if I only care about the height relative to the initial starting point, then that’s only 1 meter above. So, my final height is 1 meter with this choice.
Let’s work through the problems with these numbers. Okay. We’re ready with our list of knowns and the variable that we want. We have a lot of them written here already. We want the speed initially for this ball. As usual, the mass isn’t given. You’re going to see that, again as usual, it’s going to drop out of our equation. So, we’ve got our knowns. We’re ready now to start with our equation.
Okay. We already know that the left side is 0, that mechanical energy is conserved because gravity is the only force doing work. Wherever my speed is 0, my kinetic energy is 0 there. So, my kinetic energy final is 0. Wherever my height is 0, when I do mg times that h, it’s going to be 0. That’s my initial potential energy. So, right away, my equation gets pretty simple. I have minus ke 0 plus PE final. That’s what it simplifies to. If I add this to the other side, what I see is that the initial kinetic energy is equal to the final potential energy.
Before we plug in numbers and simplify that further, let me translate that for you. The initial kinetic energy that I had, the start of the problem, becomes or is converted to all potential energy at the end of the problem. That’s what this equation translates to and tells me about real life. Let’s put it in numbers now. That means that 1/2 mv 02, my initial kinetic energy is equal to mgh final. Well, remember I was after that initial velocity. That’s what I need to focus on. I’m going to divide both sides by m, and we see, as usual, the mass drops out. My equation then becomes 1/2 V02 equals gh final, or that my initial speed has to be the square root of 2g times h final.
When you plug in numbers, I’ve got 2 times 9.8 times 1 meter. That comes out to be 4.43 meters per second. So, let me point out that once we realized what these values were and what was appropriate to plug into our main starting equation, that what started out looking like a complicated equation got very simple quickly, and translates to a physically easy to interpret equation, which is that the initial kinetic energy all turns into potential energy in this problem. Whatever speed I had here, eventually, I pick up more speed while I lose potential, but then I lose that kinetic, and I start gaining potential. Until at the end, I have all potential. All of my original energy has now gone into potential energy.
I started out with mechanical energy, having to find this to be 0, the height here to be 0, as being all kinetic. At the end, my mechanical energy is now all potential. All of it’s been converted by the mere fact of this object being further uphill or further at a higher altitude has all been converted into potential energy.