https://youtu.be/1fhi5UHWIUM
PHYS 1101: Lecture Sixteen, Part Eight
The last example I want to do for you is the same problem that we solved in last lecture. I just want to show you that you can use this new equation, this new starting equation to solve all of these energy conservation problems. This was the problem where we had a 0.2 kilogram ball, it’s lifted upward on a string and it goes from rest to a speed of 2 meters per second over a distance of 1 meter.
Again, I’ve got rest, speed information, distance or height. I’m going to try energy conservation. In particular now, I’m going to show you that this final version or this starting equation we’ve been working with I can apply to this problem, too. This is, really will work for you in all instances.
The starting equation of course is that the work due to non-conservative forces is equal to the change in kinetic plus the change in potential energy. Let me draw again the sketch of this scenario so we be sure that we can plug into this equation properly. I know it’s the ball, it’s lifted up, so my displacement vector went in this upward direction. Here is the initial instant and here is the final instant. I start out with an initial speed of 0. I have a final speed of 2 meters per second. I’m going to call my lowest altitude, right here, the initial 0. Therefore my final altitude has to be this 1 meter. So I think that’s a pretty good sketch. Got a lot, all my information there.
So, we’re going to start out thinking about the left side of the equation, as always. In order to answer that, let me draw all the forces on the ball. Gravity is down. I have a tension force that pulls up. That’s it, there’s nothing else in contact with the ball. The work that this force does, gravity, is going to be captured by potential energy terms, but I do have another force here that’s definitely in the direction of motion. This tension’s definitely going to do work.
So my work due to non-conservative forces is going to include the work that the tension does. The work that the tension does is the size of the force times the cosine of the angle between times the distance. What’s the angle between the displacement and the tension force? These two vectors point in the same direction, they’re parallel so the angle difference is 0. That means my cosine at 0 degrees becomes 1. The work the tension does is just t times s.
Okay, your last quiz question here, question 22 is asking you if the mechanical energy is conserved. Let me walk you through a few more steps en route to solving this problem using this new equation. And then I’m going to leave it to you to arrive at the answer for question 21. We’ve thought about the left side now; let’s think about the right side. What are we going to plug into the right side of this equation?
Again, I want to think about my initial speed which was 0, my initial height which was 0, my final speed which is 2 meters per second and my final altitude which is 1 meter. Okay, my list of knowns, it’s what I have shown there. Above, in addition to, I know that the mass is 0.2 kilograms. What variable do I want? Got to be able to articulate what’s asked of me. What’s the tension of the string in terms of the specific mathematical variable that’s going to represent that? We know that that’s the magnitude of the tension force and so it’s capital T.
So now you’re ready to start using that basic equation, start plugging in to it. That means I’m going to plug in to work due to non-conservative forces equals the final kinetic minus the initial plus the final potential minus the initial potential energy. The initial kinetic is 0 and the initial potential is 0. My final kinetic though isn’t 0, neither is my final potential.
For the left side of the equation, I know I only need to include or I only include the work that the tension force does, and that I worked out above is equal to t times s and it’s positive because the tension is parallel, it’s in the same direction as the motion. So, on the left I have t times s is equal to 0.5 and final speed squared plus mg times h final.
All right, I’m going to leave it to you now to solve the rest of that problem to answer quiz question number 21. So that brings us to the end of lecture 16.