Start typing, then use the up and down arrows to select an option from the list.

Table of contents

ï»¿ >> So let's think about the problem. Here is our surface. We've got a box M, it is sliding along at some, we'll call it V1; sitting over here is a spring and that spring is attached to a brick wall. And in this region here, there is friction. And it has friction mu K. So, the box slides in, this region over here is frictionless, but as soon as it hits the spring it starts to compress the spring and it is now sliding with friction. Ok. And we can call this position X equal 0. So, let's draw the second picture where it has compressed the spring. So, here's our brick wall. Spring is now compressed it's all coiled up. Box is right against it. V2 is 0, it's come to rest, momentarily. And this is at a position X equals A. And now in the third part of this, the box bounces back off and shoots back out. So, spring comes back out to its full length, back to X equals 0. Box M takes off at V3. Ok. And the spring of course returns to its normal equilibrium position. So, let's find V final, which means we are looking for V3. And so, this is going to be position 1, this is position 2, this is position 3. So, I need to think about some principles here to attack this. What is a good principle that we could use to attack this problem? Well, how about work. Right. This whole chapter is called work. This is one of your chapter problems or at least it's very similar to one of your chapter problems. So, let's think about this from the terms of work. First off, when we are heading into the spring, what sort of energy do we have? We have kinetic energy. And then it's going to compress the spring and so we're going to do work on it, because there is friction. And we know what that friction is. It's minus FK times the distance D that it goes, which in this case is the distance A. And then it's sitting here at this compressed position. So, the energy here at E2 is spring, potential energy. Right, we have a compressed spring. And then it's going to slide back out and so in between these two we're going to do work, due to friction. And again, it's going to go across the same distance A and so it's just minus FK times A. And then it's going to shoot off and so it's going to all go back to kinetic energy. So, we come in with some kinetic energy we're going to do some work on it due to friction, we're going to have some energy in the spring, it's going to return back out, do some work on friction on the way back out, and then the box will slide along with kinetic energy. So, one way to think about how to attack this is the following: the energy in this first part is really just going to be whatever kinetic energy we have minus the work that we did. Ok. And so, we could say that kinetic energy 1 minus FKA is going to be equal to whatever energy is stored in the spring. And now going from 2 to 3, whatever energy is stored in the spring, subtract FKA and that's got to be equal to the final kinetic energy, K3. So, how do we solve this problem? This idea of doing work and looking at energy and so forth, you can sort of be careful and illustrate it like this, or you could just say the following: The energy that I start with has to be equal to the energy that I finish with as long as I keep track of it. And so, let's look at the energy that we start with. The energy that we start with is all kinetic energy. It is kinetic energy 1. And now, the energy that I finish with, is going to be kinetic energy 3 but we know that we slid across a frictional surface on the way in and on the way out, and so that is going to be thermal energy. And this is due to the frictional work. Ok. So K1 equals K3 plus whatever work I did due to friction. But thermal energy is a positive quantity and so we need to worry about the magnitude of the thermal-the magnitude of the frictional work. We know that we did FKA on the way in, FKA on the way out and so, this becomes 2FKA. And now we see what have, we have one-half MV1 squared equals one-half MV3 squared plus 2FKA. FK we're not exactly sure what that is, but we can figure it out. We know that FK is mu K times the normal force. If this whole thing is on a horizontal surface, then that normal force is just MG. So now in a problem like this, you would have mu K, you would have the mass, you would be given the input speed, you would be given how far it compresses the spring, and you would be able to solve it for V3. What's kind of interesting in this problem, which is not exactly the same as your homework problem, ok. What's interesting is, we've essentially removed the spring, right. In this equation now, we don't have springs anymore. And that's because we went from position 1 all the way to position 3. And so, in some sense, it doesn't even matter that it bounced off this spring. It could've been a rubber surface that was very springy and it just bounced directly off the wall. As long as the frictional work has the same distance, you're going to get the same answer. That's kind of cool.

Related Videos

Related Practice

03:19

09:47

04:46

05:37

04:18

05:25

05:58

04:45

© 1996–2023 Pearson All rights reserved.