A 62-kg person jumps from a window to a fire net 20.0 m direct...

Question

A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behaves like a simple spring.

(b) How much would it stretch if the person jumped from 38 m?

Accepted Answer

Welcome back. Everyone. In this problem. A group of content creators are dropping different objects on a trampoline to see what happens. They take a 53 kg dummy and drop it from the roof of a building 27 m above the ground onto the trampoline. As a result, the trampoline stretches 2.1 m from its equilibrium position. Modeling. The trampoline as a spring determine how much it would stretch from its equilibrium position if the dome we dropped from 44 m instead for our answer traces A says it's 2.1 m. B 2.5 m, C 2.7 m and D 3.4 m. Now first, let's try to visualize what's going on here. OK. So now we have a building at our content creators are dropping a Dora mm and no, for this building, it has a trampoline at the bottom. OK? So let's say that this is our trampoline and when they dropped the dummy 27 m, I dropped the door at 27 m. Then the trampoline stretches by 2.1 m. In other words, this distance is 2.1 m. No, we are saying that if the dummy were dropped from 44 m instead. Ok. So let's say in red, this is our 44 m height. Then what would be the stretch? How much would our trampoline stretch by? In other words down here, what would be our stretch X? That's what we're trying to find for or height wide of 44 m. Now, how can we figure that out? What can help us? Well, we know that based on the conservation of energy or energy at the top is going to be equal to the energy at the bottom. OK. So if we take the equilibrium position of the trampoline as the zero point, then then by the conservation of energy for the energy at the top being equal to the energy at the bottom, the potential energy initially because that would be our only energy initially here is going to be equal to the energy at the bottom. In other words, or potential energy at the bottom. OK. In this case, it would be pe B plus or elastic energy at the bottom. And we say that there's going to be potential energy at the bottom because when it stretches here that in itself is going to give it energy, it's displaced from the equilibrium point. OK? No, since that's the case, since that's the case if we were to substitute here four or 44 m height. OK, then our initial potential energy which is going to be MG multiplied by Y for, at the top is going to be equal to our potential energy at the bottom MG multiplied by Y at the bottom plus or elastic potential energy. A half of KX squared. OK. Now you notice here that we have most of the information to solve. We have the mass of the domain, we have acceleration due to gravity and we have our height yt the height at the top, we also have our height at the bottom because it's just going to be equal the X. Well, no, as a matter of fact, you know what, we don't have that height, sorry, because this is going to be the negative value of X from the equilibrium point. So here we can still solve, I mean, we can still solve uh we can still solve quadratic but we'll need to figure out the value of K. We need the spring constant before we can figure out any of this. Remember that it's the same spring constant for our previous scenario. It's the same spring constant when we dropped it from 27 m and it stretched 2.1 m. So if we can use this formula to solve four K in that scenario, then we can be able to substitute and solve eventually for X. So solving four K using the same formula then initially OK. Our mass multiplied by gravity multiplied by. Why not? No, let's call our 27 m. Why not OK is going to be equal to a mass multiplied by the gravity multiplied by negative X OK. Or a negative X not in this case, OK. Negative X not plus a half of K multiplied by X not squared. OK. Now we can use this to solve for K. Now, first of all, we know OK, that this is going to be negative, right? So if we move it over to the other side, then we'll get a half of KX squared being equal to mGy knot plus MGX knot. OK. In this case, then that means we can multiply both sides by two, divide both sides by X squared to get K to be equal to two. We can also factor out MG. So two mg multiplied by Y knot plus X knot divided by X squared or X knot squared. No, since that's the case, let's substitute and solve for K. So K is going to be equal to two multiplied by the mass 53 kg multiplied by acceleration due to gravity. We can take as 9.181 m per second squared multiplied by or height at the top. Why not? 27 m plus the height from the bottom 2.1 m? OK. All divided by 2.1 m squared. And when we substitute that into our calculator, we should find that K KK is equals to 6861.661 newtons per meter. Ok. Know that we have that know that we've solved for K and we have our directions right? Then we can substitute it to solve for our X, the X that we want to find the extension that we're really interested in. So let me come over to the right here because no, if we solve for X, well, first we can uh we can make X or make X our subject here. OK. So if that's the case, then again, we're gonna have a half of KX squared being equal to mGy from the top plus MGX. OK? And as we said, it's going to be a quadratic equation. OK. So we can solve here because it's gonna be a half of KX squared minus MGYT. OK? Plus MG multiplied by X. OK? And all of that is gonna be equal to zero. Sorry, it's gonna be minus minus uh MGX. Now, if we set it up quadratic, that means we need to get rid of the coefficient of X squared. So we can multiply the entire creation by two and divide it by K which is going to give us, OK. X squared minus two mg, divided by K X minus two MG, divided by Kyt. OK. Where YT is the height from the top of 44 m. Now let's go ahead and substitute the. So let's simplify our expression here. OK? So that's gonna be X squared minus two multiplied by 53 kg 9.81 m per second squared, all divided by 6861.66 newtons per meter X OK. Minus again this entire expression two MG divided by K I can just put it beside it here. OK? You know what I think I'll have to move the whole thing because I don't have much space now. So let me carry it down here. So it's going to be all of this. OK? Multiplied by 44 m or last term that is equal to zero. Now, we can simplify, OK. And when we do, when you simplify for our coefficient of X here and or constant, you should get X squared minus 0.15155 X minus 6.6680 being equal to zero. Now this is pretty good because now we can solve quadratic OK? Because solving quadratic that means that X OK by the quadratic formula is going to be equal to negative B. So that's negative, negative 0.15155. OK. Plus or minus the square root. Yeah, of B squared. Again, that's negative 0.15155 R squared minus four multiplied by A which is one multiplied by C 6.6680. OK. And that's all divided properly here. That's all divided by two A. So that's two multiplied by one. I know when you do that when we do that and simplify, then you should find that you get X to be either equal to 2.6591 m or equal to negative 2.50758 m. Now, since the trampoline must stretch down and it corresponds to X being positive, the trampoline would stretch to X being equal to 2.6591 m or approximately 2.7 m. So this is how much the trampoline would stretch if the dome dropped from 44 m. Instead, if we look back at our answer choices, that means X is going to be equal or, or correct answer rather is going to be c thanks for watching everyone. I hope this video helped.

Struggling with Physics?

A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behaves like a simple spring.
(b) How much would it stretch if the person jumped from 38 m?

Watch next

Simple Harmonic Motion of Vertical Springs practice set