A 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?

Hey, guys. Hopefully, you're able to solve this one on your own. But if not, here's a little bit of help. So in this first part here, we've got this spring, and it's 1 meter long, and we're supposed to stretch it out to some distance and then figure out what the spring's force constant is. So we're told that we're gonna stretch this thing out. So here's the equilibrium position, and when we're stretching it out, we're pulling some force, so we've got some applied force that's equal to 50 newtons. And then we know that that spring is gonna push back on me with the spring force, which is also equal to 50 newtons. So we're told that the spring originally, so the original length of the spring, is equal to 1.0 meters. But then once you pull it out, now it has some stretched length, which is the l final, which is equal to 1.2. So that means that this deformation here, x, is just the difference between those, which is 0.2 meters. So that's basically all we need to solve the problem. So we're just gonna set up my equation. So I know that the absolute value go ahead and put that in red. So the absolute value of the spring force, which is the absolute value of the applied force, is equal to k times x. And what we're looking for here is we're looking for the spring constant. So I know that both of these things equal 50 newtons, so that equals k. And then what's my deformation? The deformation is just the difference in spring length, so 0.2. So that means that k is equal to 50 divided by 0.2, which is 250 newtons per meter. So that is part a. Part b is a little bit different. Now instead of pulling on the spring, we're gonna push it. So we're gonna have some spring here, and we're pushing up against it, so now we're gonna compress it. So instead of stretching it out, the equilibrium position is over here, and now it's at a shorter distance. So again, originally this thing was at 1 meter, and now this new length, which is down here, this new length of final is gonna be 0.7 meters, where it still originally was 0.1. So that means that this now represents our deformation from the equilibrium, and that's just the difference between 1.7 absolute values. It's 0.3. So in this case, it's just slightly opposite. So I've got the applied force that's this way, and then I've got the spring force that's equal and opposite. And in this case, I'm actually trying to figure out what those forces are. How much force do I need to apply to push this thing and compress it to that distance? And the keyword here is we're compressing it to 0.7 meters, not by 0.7 meters. So that's the difference, so be careful with the terminology there. So let's set up our equation. We've got absolute value of spring force is absolute value of the applied force, and that's equal to k times x. Now we're actually looking for these forces. So I've got the applied force is going to be well, I need k, but I actually figured that out in the last in the last part. It's just 250. Now I just need the deformation. Well, the deformation in this is 0.3. So multiplying those two numbers, I need to apply 75 newtons of force to compress it by point 3 meters, which is compressing it 2.7 meters. So just make sure you, you know, you can distinguish those 2. Alright, guys. So let's sit for this one.

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A 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?

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