(II) The 9.0-kg mass in Fig. 8–36 is held just barely in conta...

Question

(II) The 9.0-kg mass in Fig. 8–36 is held just barely in contact with a spring for which k = 450 N/m . When that mass is released, it falls, compressing the spring and pulling the 3.0-kg mass up. How far does the 9.0-kg mass fall before momentarily coming to rest? Ignore friction in the pulley.

Accepted Answer

Welcome back. Everyone in this problem. In a storage room, a heavy box of mass 10 kg is initially at rest on a spring with a spring constant of 800 newtons per meter. The box is a part of a pulley system where it is connected to a 3 kg package. When the system is released, the heavier box compresses the spring by moving downwards, pulling the lighter package upwards, calculate the distance the box falls, compressing the spring before the entire system momentarily comes to rest for our answer choices. A says that it's 0.11 m. B 0.17 m C 0.24 m and D 0.32 m. Now let's first try to understand exactly what is happening so that we can figure out the distance the box for. OK. Now we're saying that we have our system and after it is released, the heavier box compresses the spring by moving downward. So in other words on our diagram, then our 10 kg box is going to compress the spring by a distance that we can call D OK. So let's say it compresses the spring by D yeah, the meters. Ok. And now if we think about it, the mode, the distance that it compresses the spring by is the same distance that our package is going to move upwards. So our package is also going to move upwards by a distance of D. Now, if we are talking about the system here, we know that mechanical energy is conserved. We also know that the spring follows Hook's law and we can assume that the entire system is initially at rest. So the box's fall is the sole means of energy transfer within the system based on what we know about energy. Con con con conversation. OK. Conservation. My apologies, energy conservation recall that the initial energy is going to be equal to the final energy in the system. Now let's try to picture both scenarios initially. OK. The only energy that we have in the system is the potential energy of the heavier box. OK. So let's say that initially we have ub potential energy of the box. Now after the after our heavier box is released and it compresses the spring by moving downwards. Now that the spring would probably be coiled just down to this part. OK. So you can imagine our spring just probably looks something like that then. No, OK. We know that the spring has potential elastic potential energy that we can represent as us and our package know that it's gonna be in the air. It also has potential energy So it also has, we'll also have you pee, OK. Now, what do we know about these potential energies? Well, when it comes to the package and to the box, their potential energy is going to be due to gravity. Ok. So we know that B and UK is going to be equal to the mass multiplied by the gravity multiplied by the height. And we also know that the elastic potential energy us is going to be equal to half kx squared where K is the spring constant and X is the distance by which the spring has been compressed. So applying these ideas here, that means that the, the potential energy of the box, the mass of the box multiplied by G multiplied by H in the height which in this case is D is going to be equal to the elastic potential energy of the spring, which is a half KD squared. OK, plus the potential energy of the package, which is going to be the mass of the package multiplied by gravity multiplied by distance D. Now, if you notice all three terms have D as a common factor, so we can factor OD. OK. From all three terms to be left with MBG is going to be equal to half KD plus MPG. Now this is pretty good because now we can go ahead and solve for D, we know the other quantities here, we know G the acceleration due to gravity, we know the mass of the box, we know the spring constant and we also know the mass of the crate. Now here, if we do that, let me come down to the right here. That means a half of K multiplied by D is going to be equal to MPG minus MPG. Notice both of them here are G in common. So we can rewrite that as G multiplied by MB minus MP OK. And now to get D by itself, we can multiply both sides by two and divide both sides by K. So D is going to be equal to two G multiplied by MB minus MP. OK. All divided by K. Now if we substitute, that's going to be two multiplied by 9.8 m per second squared, multiplied by the mass of the box 10 kg minus the mass of the package. What was it? 3 kg? Was it let me just double check. Yes, 3 kg. OK. And all of that is going to be divided by our spring constant K, which is 800 newtons per meter. This is how we're going to figure our distance out. Now, when we do that, if you put that expression into our calculator, you should find that you get 0.1715 m. OK. Notice that all of our answers were written to two significant figures. So if we run our distance to two significant figures, it's 0.17 m. Thus the box falls a distance of 0.17 m to compress the spring. If we look back at our answer choices, that means B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Struggling with Physics?

Watch next

Energy in Connected Objects (Systems) practice set