1

concept

## Spring Force (Hooke's Law)

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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

concept

## Work Done By Springs

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Hey guys. So up until now all of the works we've calculated had been done by constant forces like applied forces or tensions or even gravity like MG. But not all forces are gonna be constant. Some will be variable. In fact, one of the most common variable forces you'll see is the spring force which remember is governed by Hook's law. And that's just this equation we've we've seen before. So in this video, I'm gonna show you how to calculate the work that's done by springs. Let's go ahead and check this out. We're gonna come back and fill the rest of this information out. So we'll take a look at the problem here. We're pushing a box attached to a spring. We're gonna push this box to the right with some applied force. And because of that and because of action reaction, there's a spring force that pushes back on us. So we have the spring constant which is K. And we also have the deformation. We're compressing the spring and a distance X. We know that's equal to two m. Now in the first part of the problem, we want to write an expression which means we're just gonna be working with letters here. Not numbers for the work that's done by our applied force. And then the work that's done by the spring force. So how do we do this? Well if we want to calculate work then do we just use FD Cosine theta. So if you want to calculate the work done by applied force, we're just gonna use the applied force times. D. Cosine Theta here. But there's a problem with this F. A. Because remember that this F. A. From hook's law actually is equal to negative K. X. And if you can't tell from this equation what happens is this force is not going to be constant because it depends on your deformation. The more you push on the spring the more the spring force is going to push back against you. So to kind of show you how this works, I'm gonna go ahead and plot sort of a graph of the strength of the applied force versus the deformation. So what happens when you're at the equilibrium position like this? Your force is just equal to zero. And then what happens is the more you push on the spring as you push it to the right your spring force and your applied force has to increase as well. So the way that this graph looks is it kind of just looks like a straight diagonal line like this or eventually at some distance X. From just looking at letters here this F. Is just gonna be equal to K. X. So if this force is constantly changing what is the value that I actually plug into my F. A. And that's the whole point of this video guys when you have works done by constant forces you can always calculate them by using FD cosine theta. But when you have variable forces like the spring force here then we're gonna have to use the average of the force instead. So we're gonna use FD co sign data but we're gonna have to figure out the average of that force. So to figure out the work done by my applied force in a spring I'm just gonna have to figure out the average of my applied force. Then I can use FD Cosine Theta. So how do I figure out an expression for this average force? I can actually go come back and used to use this diagram here. This chart that that I've been working with here. So if my force constantly increases diagonally like this then the average is gonna be actually right in the middle. And so basically the average displacement is gonna be my X. Over two. And therefore the average force. My F. Average. If I'm going from zero to K. X. The halfway point is just gonna be one half K. X. So this is the value that I plug in for my F. A. Average. So really the work done is gonna be one half K. X. Now I have to figure out the distance. Remember the distance and the displacement are the same thing. And really this is just this X. Variable that I already have. Right? So you're pushing up against the spring some compression X. And that's also the distance that you moved this box. Right? So this is my ex as well. And then we have co sign. So what's the angle between this force and the displacement or distance? Well because my force points to the right and the distance or displacement also points to the right. Then our data is just equal to zero. So we have cosine of zero and we know that it's just evaluates to one. So basically what our expression becomes is one half K. X. Squared. So that's the first part of this. Remember that the second part of part A. Is we also want to calculate the work that's done by the spring. So this is W. F. S. And it's the same idea here. Except now we're just gonna calculate the average of the spring force and then use FD cosine Theta. But this actually we don't have to repeat this whole process again because remember that your spring force and your applied force are the same magnitude that just have opposite directions. So what this means is that the average of my spring force is also equal to one half K. X. The distance is still X. And now the cosine of the angle is gonna be different because the angle between my F. S. And my D. Is not gonna be zero degrees because remember my F. S. Points to the left and my distance points to the right. So this angle just ends up being 180. And remember this just evaluates to negative one. So really this just becomes negative one half K. X. Squared. So those are two expressions here. So basically we can see is that the work that's done by the spring when you are compressing is going to be the negative of your applied force in the same way that your hook's law, These two forces are just negative of each other and they have the same value. And so the work that's done by the spring is going to be negative one half K X squared. This is gonna be the formula that you use and your problems. All right. So let's take let's finish off this problem here and now we're actually gonna calculate plug in all the numbers. So the work that's done by our applied force is gonna be one half K X squared, which is one half times 500 times two. And this is gonna be squared. So you end up working out 2000 jewels, this makes sense. You get a positive number, basically you have to do 1000 jewels of work. You actually have to put in 1000 joules of energy in order to compress the spring. Alright. And so the work that's done by the spring force is just gonna be the opposite of this. So this is gonna be negative one half K X squared. And you're just gonna get negative 1000 jewels when you work it all out. So whatever you have for one to answer the other one is just gonna be the negative whatever number you get. Alright, so that's it for this one guys, let me know if you have any questions.

3

example

## Additional Work to Compress Spring

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Hey guys, let's take a look at this problem here. So we're told that it takes some energy to compress a spring. So I'm gonna go ahead and just draw a quick little diagram here of this spring. So basically what I have here is that when the length is one and I compress it to a length of 0.7, I'm told that that takes 200 jewels of work to do, But then I'm going to calculate how much additional work needs to be done to compress this from 0.7 To a length of 0.5. So what I'm trying to calculate here is the work. So really this is just the work that I need to do from 0.7 to 0.5 or rather that you need to do. So it's kind of tricky there, but now that we understand what we're looking for, let's go ahead and get started. The one thing I want to point out is that they were given the lengths of the spring itself, but That could be confusing because what we're actually supposed to be using is the compression distances when you're at a length of one, your compression distance is zero. When you're the length of .7, the amount that you've compressed it is 0.3. When you're at 0.5 it's 0.5. So these are the actual numbers that you need to be plugging into your equation. Not the Els. All right. So now that we got that out of the way, how do we actually go ahead and solve this? Well, basically if we want to figure out the work that's done by you and compressing from 0.7 to 0.5, we have to talk about just generally the work that's done on or by a spring between two points. If you're going from A to B, we have to slightly modify our equation for the work that is done by a spring. So we know it's one half K delta X squared. But if you're going from A to B and both of those things aren't relaxed both of those distances and the work that's done is going to be one half of K X B squared minus one half K X A squared. The reason it's B. And A even though we're going from A to B is because its final minus initial. So basically if we're trying to calculate the work that we need to do from 0. point five, We're actually gonna do is we're gonna do one half of K. And then the compression final is going to be 0.5 square. The knee compression initial says only one half the initial compression is going to be 0.3. Again, don't plug in 0.7. So you're gonna play in 0.3 squared. Now, we'll be able to just plug in all of these numbers here. Unfortunately, we actually don't know what this uh with this spring constant is. So how do we find that out? Well, there's actually one more piece of information. We know about this problem. We know that it takes 200 joules of energy to compress this spring to cm. Basically, we know this relationship right here is 200 jewels. So we can do is we can say the work that we do in compressing from uh 20.7 is actually just gonna be one half of K. X. Square. So we're going to do here is we're going to say that this is equal to 200 jewels and we're compressing it from a relaxed distance. so there's no other term that needs to go here. So we have one half of K. And then our initial compression distance is 0.3 square. This is equal to 200 jewels. And if you go ahead and work this out, what you're gonna get for K is you're gonna get 4444 newtons per meter. So this is sort of the missing piece of the puzzle that we need now that we've figured out the K. Constant. We could just plug it back into this expression here to figure out the work done the additional work that we need to do. All right. So if you go ahead and work this out for you to get, is that the work that's done by you from 0. point 70.5 is going to be one half of times the X final, which is 0.5 squared minus the one half Of 4,444 times the X initial, which is 0.3. Go ahead and plug it in. What you're gonna get. Is that the work that's done is equal to 356 jewels. Notice how we get a positive number. And that makes sense because we are actually compressing the spring. So we need to do some additional positive work to compress it even further. All right. So notice how also it's not linear. From 0 to 0.3. We actually needed only 200 jewels of work to do this. Now we're compressing it only an additional 0.2. And it actually takes more work. So from here to here, it's actually going to take 356 jewels to compress it. So that has to do with the one half K X squared. Right? So it gets harder and harder and harder. The more you compress the spring, you have to do more work to it. All right, so that's it for this one. Guys, let me know if you have any questions.

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