(III) An engineer is designing a spring to be placed at the bo...

Question

(III) An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable breaks when the elevator is at a height h above the top of the spring, calculate the value that the spring constant k should have so that passengers undergo an acceleration of no more than 4.5 g when being brought to rest. Let M be the total mass of the elevator and passengers.

Accepted Answer

Welcome back, everyone in this problem. An adventure sports company is installing a safety net with a spring mechanism below a bungee jumping platform to provide an additional layer of safety, the net needs to be designed so that if a jumper's bungee card were to fail at the lowest point of their jump, the spring in the safety net would decelerate the jumper to a stop with an acceleration no greater than three G. Assuming the lowest point of the jump is at height edge above the safety net, calculate the spring constant care necessary for the safety mechanism. Let envy the mass of an average jumper for our answer choices. A says that the spring constant is nine MG divided by two HB six MG divided by HCMG, divided by H and DMG divided by two H. Now, first, let's try to visualize what's going on here. So we're talking about a jumper, OK, who's connected to a bungee cord and in case he falls, there's a safety net to catch him and the jumper is eight units above the safety net. In the safety net. It has a spring that just in case our jumper falls. OK? Just in case our jumper falls, then it will decelerate the jumper with an acceleration of three G. OK. So here we know that that acceleration is three G. We know that the, the weight of a jumper is going to be his mass M multiplied by G. OK. And now, from this problem, we are trying to figure out the spring constant K. So let's ask ourselves, is there anything we know here that can be related to our spring constant in our mechanism? Well, if we think about it, OK, assuming that the spring behaves linearly within its elastic limit and air resistance is negligible during the fall of the jumper. Then the potential energy of the jumper just as he is about to fall is going to be equal to the potential energy that's stored in the spring given that he requires the maximum deceleration of three G to stop. OK. So in this case, by the conservation of energy principle, the potential energy of the jumper, that's the potential energy due to gravity is going to be equal to the potential energy that's stored in the spring. Now, what do we know about both those potential energies? We recall, OK, that the potential energy due to gravity is equal to mass multiplied by, let me write that out mass multiplied by gravity are the acceleration due to gravity multiplied by the height. OK. And the elastic elastic potential energy in the spring is equal to a half of KX squared, where K is the spring constant and X is the compression of the spring. In this case, our maximum compression. So since that's the case, I think I could probably show that a bit clearer here. So lets say that is this a here. So since that's the case, OK, since that's the case, then applying it to our problem, that means the mass of the jumper M multiplied by GH is going to be equal to a half of KX squared. Now, we're almost on our way because this just tells us all the potential energy of the jumper is converted into spring, potential energy. But we can't solve for K because we don't know X notice that when we look on, on all of our answers, none of them have X in there. None of them uh include the vari for the compression of the spring. They are all expressed in terms of MG and H. So if we can rewrite X in terms of some of those variables and even K, then we should be able to figure out what the value of K is going to be. So let's think a bit more about the compression of the spring. Now, what else do we know here that can help us? We're talking about the jumper falling into this distance here. OK. Well, for our spring, we know that it's going to exert a force on the jumper that's proportional to the length of our compression by Hook's law. OK. So by Hook's law, OK. We call that by Hook's law, our force that the spring will exert is going to be equal to the spring constant K multiplied by the compression of the spring X. We also know that by Newton's Second Law, OK, that force is going to be proportional to the acceleration by the formula that says F equals ma. Now, with that in mind, since we're talking about the same force here, that means we should be able to equate both of these equations to get an expression for K. Sorry for X because recall that we already know that acceleration is going to be equal to three G. So by that idea, then that means KX is going to be equal to M the mass of the jumper multiplied by the acceleration. The amount of deceleration the spring can give which is three G. OK. Which then means that X is going to be equal to three mg, three mg divided by K. Now that we have an expression for X, we can substitute it into our formula that dealt with the energies cause now putting that back in red. Therefore, that means MGH is going to be equal to half of K multiplied by X which is three MG divided by K all squared. Now, if we continue here, that's going to be a half of K multiplied by three squared, nine M squared is M squared. G is going to be squared and K is also going to be squared. OK. And now both of those are multiplying notice here that we can simplify, OK. K goes into K squared K times, right? And here M goes into M squared M times, we can cancel M from both sides. Likewise, we can cancel G from both sides and G into G squared goes G times. So this tells us then that on our left hand side, we'll be left with each and on our right hand side, we're going to have a half. OK? Of nine mg divided by K. Remember we're solving for K. So to do that, we can simply multiply both sides by K. So now that tells us then that care. Yeah. Well, not simply multiply both sides by K, but also divide my apologies, both sides by H. So when we do that H Cancers H on the left hand side, that just leaves us with K and on the right hand side, the K will the keys will cancer leaving us with nine mg divided by two H. OK. So nine MG divided by two H. Thus, we can use this expression to find the spring constant. And when we look back on our answer choices, that means A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Hooke's Law

Newton's Second Law of Motion

Acceleration due to Gravity

Watch next