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Hooke's LawWhat is the mass of catfish that stretches spring scale 6.25 cm? The force constant of the _ scale 5 mechanism is 1.25x103 Nlm.7.96k85.6kg78.1N78.1k...

Question

Hooke's LawWhat is the mass of catfish that stretches spring scale 6.25 cm? The force constant of the _ scale 5 mechanism is 1.25x103 Nlm.7.96k85.6kg78.1N78.1kB2k8

Hooke's Law What is the mass of catfish that stretches spring scale 6.25 cm? The force constant of the _ scale 5 mechanism is 1.25x103 Nlm. 7.96k8 5.6kg 78.1N 78.1kB 2k8



Answers

A freshly caught catfish is placed on a spring scale, and it oscillates up and down with a period of 0.204 s. If the spring constant
of the scale is $2160 \mathrm{N} / \mathrm{m},$ what is the mass of the cattish?

In this example will be looking at a vertical mass spring system in which we have a scale basically that we use to measure fish. And the way it works is we have some spring, the rest at an equilibrium position. Um, when it's just hanging down vertically and whenever it stretches, we measure how far down it stretches from its equilibrium position. So these could be like measurement marks here and based on how far down it stretches, well, tell us how far or how much the fish ways. So the first thing we want to figure out is what our spring constant is. And so the way we're gonna do this is we need to have a given mass and displacement, and then we can basically calibrate this scale by finding the spring constant associated with it. And the trip to do this is to realize that when the scale is it rest here that we have a situation in which are fishes and static equilibrium and the sum of all forces in the Y direction is equal to zero. And those forces are just going to be the um spring force that's pulling it back up in this direction, which I'm actually gonna call minus why? And then we'll have the force of gravity pointing in the downward direction, which I'm gonna let be positive. So we have an M g and A minus K X for our forces here. So what this means is that at static equilibrium, we will have mg equal to K X. If we add text either side and then we can just sell for K X. And we've got a way to find our spring constant. Okay, so now ah, we have to do is just plug in our displacement, which is Delta. While it's kind of getting wild here and putting X isn't instead so let's just remedy That was real quick over here. That one's OK to do this one. All right, so we have our displacement, right, cause I'm going to head. Go ahead and say that our equilibrium position is why equals zero just for the sake of convenience. So if we plug in all of our values there, we will get that are spring constant K is equal to 1200 and 26 newtons per meter. Okay, so now that we have this spring, constant for this system, we will be able to determine the displacement that, if it should have given its mass or the massive fish should have given how far it displaces. And that's what the other parts are gonna be about. So, Barbie, it's gonna be given a displacement of 5.5 centimeters. Tell us what the mass of the fish must be. So now we're using this like a scale, and we're in the exact same scenario where again we have a balance of the force of gravity and the force from the spring being displaced, and now we're just holding for em. So we'll divide both sides by G. And since we have this spring constant now we can just go ahead and plug everything in. And, of course, in the this previous example I converted are why here, two meters by moving the place two times back. So this was 0.8 meters and I'm gonna do the exact same thing over here. Uh uh, This X equals 5.5 centimeters, so this would be 0.55 centimeters. And again, I'm kind of, uh, not being very considerate about my exes. and wise here. But I promise you, they mean the same thing. Okay, so now we have our acts. We have the spring constant and gravity. So we just put all our values in and we'll get it. Mess. Ah, 6.874 kilograms. Okay, so now for the last part, we're gonna ask another question about our scale here. And the question is, how far apart are the half kilogram marks? So you can imagine there We have a little tape measure next to our, um, scale here, Right? And maybe each of these little ticks here, it's supposed to rent represent Ah, 1/2 kilograms worth of distance. So we want to know well, the actual distance of each of those. So how far will are ah, scale extend if we add another 0.5 kilograms on top of it. So we're going to do the exact same scenario. And really, all I'm gonna do is just imagine now that we start our scale off at equilibrium, so it has nothing on it. And then we go ahead and hang a 0.5 kilogram mass from it. And the distance that it stretches is going to be the distance for this particular spring that it will stretch given that we place 0.5 more kilograms on it. So this problem is pretty much identical to ours. Were again. We have some static equilibrium. Situation with the force of gravity is matching the force from our stretch spring. And this time we're just solving for the displacement and our mass is gonna be 0.5 kilograms. Still divide both sides of this equation back a and again, this is the same springs. Okay, has not changed. G never changes, luckily, and we've got em. So again we could just plug everything in and we will find that the distance between 1/2 um kilogram marks, it's going to be 0.4 meters or we've been doing everything in centimeters, so we could say zero point four centimeters. So we could make a pretty handy, um, meter next to our scale here, where instead of putting distances on it, right, we could go ahead and put weights on it so that we could easily read, just read off the meter toward the way to the fishes.

In this particle. Their case. We're going to use hooks. The ethical SK X, where f is equal, do MGI So from here we can say that gay equals mg over X. We're using massacres 500 gram or 0.5 kilogram G's 9.8 meters per second squared, divided by the displacement is two centimeters. So that is 0.2 major and the unit off. This was the new Don't automated. So that is 4.9 divided by point or do This is 245 new done far meter that is answer for part A For part B, we are once again going to use the same form. The M G equals K X, but now you're trying to find m the mass off the fish, which is X over G case found to be 245. The fish displaces the spring by 4.5 centimeters. Appoint all 45 meter divided by 9.8. That will come out to be point 1.1 to 5 kilograms

In this party killer case we're going to use force equals K X, which is the hooks all but for us wait that he will put again forces MGI So from here we will be using M G equals kicks for the first part off the problem. We're trying to find the spring constant of the spring K, which will come out to be mg over X from this formula. So mass off. The spring mass of the fish is 500 grams. So that is 0.5. You know Graham G's 9.8 meter per second squared, divided by the displacement is two centimeter are 0.2 major, so that will come out to be 245 Newton are major for part B. You have to find what is the mass, and we're going to once again use the same formula Where Massey's okay, x over g key. We just found it to be 245. You don't buy meter. The displacement of this being is for born five centimeter or the 4.5 centimeter. We're going to write guest 0.0 for five meter divided by 9.8 meters, far second square, all together pun ticket. In the calculator, it comes out to be 1.1 to 5 kilogram.

So to find the value off spring costume K, we are going to use hooks Law F is equal to negative cakes. In this case, we don't care about this negative sign because all we need in this case is the magnitude on we arguing that fish off 10 kilogram is attached to the spring. So we are going to equate host you to gravity to K X. And from this we can find K s m j divided by ex. So I assumed that the spring is like this on Lord is attached to this hook here which weighs about 10 kilograms. So this will be the fools acting downwards. And this will be the fools acting upwards which is for this course. This is gravitational force. This is springs force. So let's substitute all the values capable to 10 times 9.8 divided by the displacement which is eight centimeters, 0.8 meters c. I kept value in meters so answer here is 1 to 25 Newton who meter in part B. We are asked how much master will stretch the string. Bye bye point Fight again. We will use the same formula omg is equal to K X and in this case we are given X being okay. We are supposed to find the value of mass. So mass is equal to 12 to 5 time still value effects, which is 5.5 centimeters, 0.55055 divided by 9.8. So if you do this, you will get six point 875 kills on in the party. We are asked how How far apart are half kilogram marks on the scale? So let's find out X for half kilogram. So we know masses half kilogram times 9.8, divided by 12 to fight. This gives you the answer. Zero points use you for Meet us, I think.


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