17. Periodic Motion

Spring Force (Hooke's Law)

1

concept

## Spring Force (Hooke's Law)

5m

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So whenever you're pushing or pulling against the spring with some applied force, that spring pushes back on you in the opposite direction and it's because it forms an action reaction pair. So for instance, in this example, I'm gonna take this spring and I'm gonna push against it with some applied force, that's the action. And then the spring pushes back on me with an equal but opposite direction force, that's the reaction force. And so we say that the spring force is equal to the negative of your applied force and that's equal to negative K times X where X. Represents the objects deformation here and then K. Is just a constant. Just a force constant. So the only thing you need to know about this negative sign here, that negative sign just means that that force points in the opposite direction is all it means. And because that that negative sign often just goes away. And we're only just asked for the magnitude of these forces. We're gonna use this more powerful equation that the absolute value of all these forces are equal and that's just equal to K times X. The negative sign just goes away. So to see how all this stuff works, let's just go ahead and take a look at an example. So I've gotten this first example here, I'm pushing this spring to the left. So that means that my applied force is equal to 1 20. And so that means that the spring force is in the opposite direction, also equal to 1 20. Now I'm told that the K constant, which is that force constant is equal to 20. And what that K constant really just means is it's a measure of how stiff the spring is, how hard it is to push it or pull it. And so what I'm asked for is how much it compresses by. So I've got these two forces and I'm pushing on this spring and I've compressed it by some distance. X. And that is the objects deformation. That deformation is always measured relative to this dotted line here, which is basically the relaxed position. That's right here. So that is the relaxed position. By the way. The other word that you might hear for that often is the equilibrium position. So equilibrium and that's where X. Is equal to zero. Okay, so let's go ahead and just set up my equation here. So I've got the magnitude of these forces are equal to each other and that's just equal to K times X. I'm looking for X. And I've got K right and I also got the force. So my applied force, which is the springs force was just equal to 1 20. My K. Is equal to 20. And then I've got X. Is that is my variable here. So you go ahead and solve for that and you get a deformation of six m. That's how much I compressed it by. So let's take a look at a different example here because here I was pushing against the spring. Now what I'm gonna do is I'm gonna pull against it. So now what happens is I'm taking this spring, I'm pulling it out. So my applied forces here, which means that the spring force is equal and opposite to that. But now I'm actually looking for how much force I need to pull it out to some distance here. So let's take a look. I've also got that the K constant is equal to 40 in this equation. So let's just write out my formula F. S. Equals F. A. Equals K. Times X. Now, I'm looking for the actual forces here, I need I know what the K. Is. Now, I just need to know what my deformation is. So let's take a look at these numbers. I've got 10 m 16 m. Which one of those things represents my distance. My deformation. Well, before we actually had a compression distance that was given to us. But here, all we're told is that the spring was originally 10 m. And then once you've pulled on it, the spring has now become 16 m long. So really this distance in between here, the amount that you've pulled on it or reformed it, is that X. Right, so that's equal to six m. So what that means is that the X. Is not about the springs length. The X. Is actually the difference in the springs length from the final to the initial. So now I've got my X. That's just equal to six m now, Just go ahead and solve the problem. So I've got F. S. Is equal to uh So yeah, so I've got 40 and then I've got six here. So I've got the spring force is equal to 240 newtons. So I've got that. So let's take a look at what happened here. Right. And this first example, I had an ex or deformation of six. I had a K constant of 20 and then the force was equal to 120. Well in this example I've got the same exact deformation six. I've got the K constant with 42 double that. And the restoring force. Or sorry, the spring force was equal to 240. So what happens is I doubled the spring constant, the force constant and then I doubled the force. So you can actually see that. Just just using the equation itself, F. S. Equals K. X. So if the X. Is the same. And then this thing gets doubled, then then the force has to be doubled as well. So what that means is that this K. Represents how hard it is To deform a spring, you have to use more force to deform it the same exact amount. And so the units for that, the units for K are going to be in newtons per meter. How much newton's you have to do per every meter of displacement. Now, you should remember that. But if you ever forget, it's just F. S. Equals K. X. You can get to that. What are the units for F. S. That's just in newtons. And then you've got K. And then the units for X are in meters, so you just move it over to the other side and figure that out. Okay, so in both of these examples here, right, Whether I've pushed up against the spring or I've pulled against it, the force that acts in the opposite direction always wants to pull the system back to the center. And so what that means is that F. S. Is a restoring force. We call that restoring force, and it always opposes the deformation. It always opposes your push or your pull. So that's everything we need to know about Hook's Law. Let's go ahead. And you do use some examples.

2

Problem

ProblemA 1.0 m-long spring is laid horizontally with one of its ends fixed. When you pull on it with 50 N, it stretches to 1.2 m. (a) What is the spring's force constant (b) How much force is needed to compress it to 0.7 m?

A

250 N/m; 175 N

B

50 N/m; 15 N

C

250 N/m; 75 N

D

50 N/m; 35 N

3

concept

## Acceleration of Mass-Spring Systems

3m

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if you attach a mass to a spring it becomes a mass spring system. So now you're not pushing up against the spring itself, you're gonna take some block and push it up against the spring with some applied force. Now we know that that spring is going to push back against you in the opposite direction with some F. S. And we know that those two forces are equal to each other except for this minus sign here. So you say that F. S. Is equal to negative F. A. And that's equal to negative K. Times X. So that means that you're pushing up against something and you're compressing at some distance here. And so now if we consider all the forces that are acting on this object, we can use F. Equals M. A. To figure out what's going on. So that the first thing is that this m always refers to the mass of the object itself. And so we're always going to assume that the mass of the spring is equal to zero. So now we write all the forces, we've got negative F. A. And then we've got the positive spring force. And those two things are equal and opposite. So they're going to cancel out. So that means M. A. Is equal to zero. And so therefore if you're just pushing up against this thing and keeping it there there's no acceleration. So now what happens if I release that applied force? So if I remove my hand now the only thing that's pushing up against the spring or this object here is that spring force the applied force goes away. And so that's equal to negative K. X. So now the spring force is the is this force that's gonna want to push it or pull it back to the equilibrium. So now if we consider all these forces here we've got Fs equals M. A. Now we know that's K. X. So we have negative K. X. Whoops, negative K. X. Is equal to mass times acceleration. This is a really really really powerful formula. And so now if we want to solve and calculate for the acceleration we can just go ahead and divide over the mass and we get acceleration is equal to negative K. Over M times X. Where again this negative sign just reminds you that it's in the opposite direction of whatever you're pushing or pulling it. So let's check out an example. So this example here we've got a 0.60 kg block that's attached to some spring here we've got the cake constant is equal to 15. Let me move that somewhere else. And we're told that this thing is stretched .2 m To the right beyond its equilibrium point. So we've got this deformation X. is equal to 0.2 m. So now given those two things, we're supposed to figure out the force that's acting on this object and also its acceleration. So in this we're being asked for the force on the block. So that's gonna be the spring force. So let's write the whole equation out. We've got negative K. X. And that's equal to N times A. So if I wanted to figure out the spring force here, all I need is the compression distance or the the stretching distance and the force constant. And I have both of those. So it's going to be negative 15, that's my spring constant and then 0.2 for the deformation. So I've got the spring force is equal to negative three newtons. Now, the reason we got a negative sign is because we're taking the right direction to be positive. So this negative sign just means it points to the left. That makes sense because once you release it, that force is going to be acting in that direction. So now we're supposed to find the acceleration. So let's just go ahead and use our formula, we got an acceleration formula A is equal to negative K over M times X. We've got all of those numbers. So A is just equal to negative 15, divided by 0.6 times 0.2 And we get an acceleration that's equal to negative five meters per second squared. Again, that negative sign just means it points to the left. That makes sense because if this is the only force that's acting on this thing, that means the acceleration must be towards the left. If you ever forget this formula here, you can always get back to this just by using the spring force is equal to mass times acceleration. Those two things are equal to each other. Alright guys, that's it for this one. Let's keep.

4

Problem

ProblemYou push a 3-kg mass against a spring and release it from rest. Its maximum acceleration is 10m/s<sup2></sup> when pushed back 0.5m. What is the (a)spring constant and (b) restoring force at this point?

A

60 N/m; 30 N

B

120 N/m; 45 N

C

100 N/m; 35 N

D

30 N/m; 15 N

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