https://youtu.be/M3eqGlNQelk
PHYS 1101: Lecture Sixteen, Part Five
All right, here’s our first example. A cyclist approaches the bottom a gradual hill at a speed of 18 meters per second. The hill is 4.2 meters high and the cyclist estimates that she is going fast enough to coast up and over this hill without pedaling. We get to ignore air resistance and friction, don’t have to worry about any work that these forces might do, and we have to determine the speed at which the cyclist crests the hill.
I’m going to apply energy conservation to this problem because I see here a comparison of speeds and position. It’s the location when she crests the hill compared to when she’s at the bottom of the hill. That’s the wording there that lets me know that energy conservation is probably a useful tool to throw at this problem.
Here’s my starting equation for that tool. Work due to the forces other than gravity is equal to kinetic energy final minus kinetic energy initial plus potential energy final minus potential energy initial. To digest what these terms are, what needs to be substituted in, I’m going to draw myself a quick sketch so I don’t get confused. Here’s my bicyclist, that’s excellent, that’s her, at the start, she’s at the bottom of the hill.
The end of the problem is with her cresting the top of the hill, I’m going to sketch her there. And, here’s my object, let me clearly label that this is going to be my initial instant, we’ll have an initial height and an initial speed. This is my final instant, with my final height, and my final speed. The lowest altitude in the problem I’m going to define as 0 altitude, make my 0 there.
Okay, to think about the left side of this equation, I need to picture all the forces that are on this object and I’m going to have to include the work done by everything except gravity here. Let me draw the forces on this cyclist. So, anywhere in between, let me draw her here, anywhere, from start to finish, I’m know that I have her weight, and I have a normal force. I don’t have to worry about friction, air resistance, nothing else is touching the bicycle that’s all we have. Okay, the normal force can never do work, because it’s always perpendicular to my displacement. Let me move this down a little bit.
All right, the surface always pushes perpendicular to it, pushes away from the surface. If I’m always moving along the surface, then this is always going to be 90 degrees. Normal forces never do work. The only other force I have there would be work represented by gravity and that’s going to be captured on the right side of my equation with my potential energy terms here.
So, I’m not going to, I don’t calculate gravity’s work and put it on the left. I’ll catch it by doing the potential energy terms. So, w due to non-conservative forces then, is going to be 0. The only force I have doing any work is gravity, and that’s not included there. So, is mechanical energy conserved, yes. Left side of my equation is 0; the change in mechanical energy, then, is 0.
Okay, now let’s think about the right-hand side. For the right hand side I’m going to need to substitute in kinetic energy at the two instances and potential energy at the two instances. So, to prepare myself to do that, let me just think about what the speeds are at the two instances, do I know those values, and what are the heights.
So, let me put here, right side, what’s my initial speed, do I know it? Yeah, at the bottom of the hill, she’s going 18 meters per second. What’s her altitude at that position? Well that’s where I’ve called my 0. How about at the top of the hill, the final instant, do we know the speed? No, in fact, that’s what we want to find. We want to find the speed at which the cyclist crests the hill. Do we know the height, her altitude at the top of the hill? Yeah we do, it’s 4.2 meters high. So, I’m prepared now, to plug into the right side.
Before we start working with our basic equation, let’s see if we can add anything more to our list of knowns. We really can’t, I think we’ve pretty thoroughly done that as we’ve thought through the variables and the values we needed for the right side of the equation. What variable to we want? We want the Vf, that’s the variable that will show up in our equation, that’s going to represent her speed at this instant.
So, I’m ready, now I’m going to start with my basic equation, and start plugging in for it. Okay, we concluded that the left side was 0, mechanical energy is conserved. I’m going to, just, write out for this first time very carefully what each term is. Sorry about that. Oops, for the potential energy now, I’m going to substitute in, it’s mghf minus mgh initial. Okay, a bunch of these are 0. Well, the initial height is 0, this term is 0, this is 4.2 meters, I know g, I know, do I know m, I didn’t write that in my list of knowns, I don’t know m. I know V0, and this is what I want, the final speed.
So, other than not knowing M, this looks like I’m in good shape. Let’s do some simplification before we panic. Do you recognize that you can factor M out, that it’s a factor common to all of these terms. When you do that you end up with 0.5 Vf squared minus0.5 V0 squared plus gh final.
Okay, now that I’ve factored it out and I see that it’s just a factor, I realize that it does just drop out, because I can just imagine dividing both sides by M, and I still end up with 0 on the left hand side. Okay, so we write in that I still have 0 on the left side is equal to 0.5 Vf squared minus 0.5 V0 squared plus GH final. When I do my algebra and solve for the variable I want, I end up with that Vf is going to be the square root of V0 squared minus 2gh final.
And, here are those numbers plugged in, here is V0 squared minus 2 times g times the final height, my final altitude, given that the bottom of the hill is 0. And, when I plug those numbers into my calculator I end up with 15.5, and the units are going to be meters per second. I can see that here I’m going to have meters squared per second squared for both of these. When I square root it, that leaves me with just meters per second.