Physics Problems: Mechanical Energy Conservation

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Conservation of mechanical energy - it's a really deep and important concept in physics and it's a great problem-solving device. Conservation of mechanical energy says change in kinetic energy plus change in potential energy is equal to 0. There is no change in the sum of kinetic energy plus potential energy. Another way to write that, that there's no change, is to say that the mechanical energy, kinetic energy plus potential energy is a constant. It's never changing, it's the same at the beginning of the problem and at the end of the problem. Yet another way to write this down is E[mechanical] at the beginning is equal to E[mechanical ]at the end.
So this formula looks deceptively simple and it does have some limitations. For instance, it's only true for systems where this describes essentially all the energy of the system. The kinetic energy describes energy of motion, potential energy tells you how much energy you have by virtue of where you are. So, for instance, gravitational potential energy mgy is just a function of how high you are. For a spring the potential energy is a function of stretched it is. There are times in life when this formula is not true. A generalization of this formula is change in kinetic energy plus change in potential energy is work done by other forces, other non-conservative forces. For instance, perhaps there's friction in the story and friction can do work. It's negative work when there's friction, and so the kinetic energy plus the potential energy will be changing if there's friction; they will be decreasing. The other work might also be may be me; I'm pushing on something. There's no potential energy associated with me, I'm just doing some outside work on the problem and changing the energy of the system. But you can usually recognize problems where mechanical energy is conserved. It's problems where there are outsiders. You can see all of the forces and all of the forces are conservative forces, and then mechanical energy conservation is quick and easy to use.
You know, mechanical energy conservation tells you some very simple things. If you take a ball and you drop it - let met just start it and drop it and it falls down, hits the bottom, comes back up. Let's do it again. I dropped it, it fell down, it's energy was converting. It was all gravitational potential energy, and then it bounced off the bottom and, at that moment in time, its energy was all kinetic energy, and then on its way back up again its kinetic energy was turning back into gravitational potential energy. And, at the end, it got back basically to where it started, a little bit lower. mgh[final] was little bit less, because there is some loss of energy when you hit the bottom, and there's also a little bit of air drag. But it's a pretty decent approximation that mechanical energy is conserved throughout the whole process, not just at the top, but at any point in the process.
This is a case where we don't need conservation of mechanical energy, I just used Newton's Second Law. It's simple motion, just gravity, but what if you have a complicated problem with a both of different forces and maybe Newton's Second Law starts to look a little bit difficult. Here's a classic example, the roller coaster problem. You start off at rest at the top of a roller coaster of height h[initial]. You've got a complicated shape, it does a loop the loop, and you end up here at the end at some final height, h[final]. There are lots of questions that I could ask about the roller coaster as it travels along, and some of those questions I cannot answer with conservation of energy, like, "How long does it take?" Well, that's a tough question, but here's one that I could ask that you might be very interested in both as a rider and as a person designing the roller coaster: "How fast are you going when you get to the end?" It's a perfectly legitimate physics question. We know our initial conditions, I'd like to know the final conditions. And do I have to know all the fancy details about the shape and everything? If you're thinking Newton's Second Law, which is a perfectly legitimate way to think about answering such a question, you start getting uncomfortable, because, as the thing is moving along, you have to draw force a diagram at every spot along the track. And, at every spot along the track, there's always gravitational force down and there's always a normal force, which is perpendicular to the track, so it's changing in direction and in magnitude as you cruise along. And that's going to be a hard problem to solve. I'll guarantee you that the net force will not be constant, which means that the acceleration of the car will not be a constant, so we can use any of our constant acceleration kinematic equations. You could, in principle, solve this problem with a computer that takes acceleration and integrates acceleration as a function of time defined velocity as a function of time. What a nightmare! It'd be an awfully hard problem. I wouldn't even want to tackle it.
But with conservation of mechanical energy, I'm really all set up. It's an easy problem, because kinetic energy plus potential energy is never changing. And I know exactly what it is at the beginning of the story. Let's compute it symbolically. In the beginning, we have kinetic energy is 0 - I started from rest, and the potential energy is mgh[initial], and that's it. That's the total initial energy, the total mechanical energy at the start. Now, of course, I'm assuming there's no other forces, no friction, in particular, in which case the initial energy should equal to the final energy. So what is the final energy when it reaches the end of the track? Well, it's moving, I presume, mv[final ]squared. And it's not at ground level. Remember, it ended at a height h final, so its potential energy at the end is not 0, it's mgh[final]. There's a nice formula. Interesting fact - it doesn't matter what the mass is. That's great for roller coaster designers. It doesn't matter how many people you pack into the car, the final feed is going to be given just by the solution to this equation. If you solve this, v[final]is equal to the square root of - multiply through by 2, there is a g and I get h[initial] minus h[final]. Stare at this formula for a second. h[initial] minus h[final], and that's a positive number if you start higher than you ended, and that's good, because you want a positive number under the square root. And you could plug-in some numbers. A big roller coaster might start off 55 meters above the ground level. That's pretty high, that's over 150 feet, a 15-story tall roller coaster. And maybe it doesn't quite end at ground level, maybe it's 5 meters above ground level, so 55 minus 5 would be 50-meter drop. Plug in g and multiply it out, the answer is about 30 meters per second, or about 60 miles per hour. So you're really cranking at the end of this roller coaster ride, and not at a completely unreasonable number. Easy to solve this problem, using conservation of energy. I would really hate to have to solve this with Newton's Second Law.
I can solve even more complicated problems. Let me take that roller coaster story, so I'm a roller coaster designer, and I'm thinking we're going 60 miles per hour, but it's the end of the line. I've got to get people out of the train. What am I going to do? Well, here's a neat idea, I've never seen this on a roller coaster, but let me take a big old giant spring and mount it on against a big wall, and I'll just put that spring right here. So the roller coaster is approaching the spring, and now we know at what speed, and it's going to squish the spring and, at a certain point, because springs supply steadily larger and larger force, it'll come to a halt. And then you could have a little ratchet go up and stop the cart and everybody gets out. If you go and buy the spring, somebody tells you what the spring constant is - if they don't tell you, you can measure it, because f equals kx minus kx, so you just push on it with a known force and measure how much it compresses. So k would be something that you would know when you buy this spring. And what I would like to know is where do I set the latch? What's the stretch going to be? Now, when you use conservation of energy, you might think, "I've got to do this as a two-step problem." First, what I just did, use conservation of energy to figure out the speed, and then I could use conservation of energy a second time. The energy at this moment -that's when it's still moving, just before it hits the spring, is equal to the very final energy. And what's the very final energy? What's E[final ]now? Well, it's still at height hf and it's come to a halt, so it's got zero kinetic energy and you still have mghf of gravitational potential energy, and now we have to add in another potential energy, because now there's a spring in the problem, kx squared. And the only unknown is x, how much does it compress, and you could simply solve for it. You don't have to do it as a two-step problem, because E[f] is equal to E[initial], it's equal to mgh[initial]. And so you don't even have to know what the speed was just before it hit the spring. It's a nice thing about conservation of energy, you can relate the energy at any stage of this motion to the energy at any other stage. So there's a complicated-looking problem with springs and roller coasters and a solution is really straightforward.
Let's do one more problem, supposing that you got this roller coaster set up and they haven't mounted the spring yet, so the track just ends and, oops, somebody launched the roller coaster. So it's cruising along, gets to the end and, at the end, it goes flying, people screaming, horrible story. What's going to happen? Well, I would like to know how fast are the people going just before they hit the ground, to see whether they're going to survive or not. I want to know the final speed, just before they hit the ground. So now you're thinking, "Uh oh, I know how I have to do this." Right after you leave the track, we're no longer a roller coaster problem, it's a free fall problem. It's one of those two-dimensional kinematics problems, where you know the initial speed and the initial angle of launch and you have to solve the equations of constant acceleration kinematics in the x direction to find v[final ]x and in the y direction with g as the acceleration. It's this kinematics of a ballistic motion. But no, you don't have to do any of that stuff; you just use conservation of energy. At the beginning, you had mgh[initial] gravitational potential energy and zero kinetic energy, and then, at the end, you're at ground level. You've got mv squared of kinetic energy and you're at the ground level, so there's zero final potential energy. Mass cancels again, you solve for the final speed, it's just the square root of 2g h[initial]. In fact, if you plug-in 55 meters, it's not all that much bigger and it's going to be little bit more than 60 miles per hour.
So you can solve these problems. Some of them we could solve before, but they were hard, some of them we couldn't solve before at all, and we've got this nice straightforward method now, conservation of mechanical energy. It's a new conceptual way of thinking about problems and it's a really convenient way of tackling a lot of physics problems.
Energy
Conservation of Energy
Solving Problems Using Conservation of Mechanical Energy Page [2 of 2]