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

Additional Work to Compress Spring

Patrick Ford
Was this helpful?
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.