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Two blocks; of masses M 2.80 kg and 2M, are con- nected to 3 spring of spring constant k=230 Nlm that has one end fixed a5 shown in Fig; 8-25. The hori- zontal su...

Question

Two blocks; of masses M 2.80 kg and 2M, are con- nected to 3 spring of spring constant k=230 Nlm that has one end fixed a5 shown in Fig; 8-25. The hori- zontal surface and the pulley are frictionless, and thc pullcy has ncg- ligible mass. The blocks are released from rest with the spring relaxed: Figure 8-25, Problem 21_ (a) What is the combined kinetic energy of thc two blocks when the hanging block has fallen 0.120 m? (b) What is the kinetic energy of the hanging block when it has fallen t

Two blocks; of masses M 2.80 kg and 2M, are con- nected to 3 spring of spring constant k=230 Nlm that has one end fixed a5 shown in Fig; 8-25. The hori- zontal surface and the pulley are frictionless, and thc pullcy has ncg- ligible mass. The blocks are released from rest with the spring relaxed: Figure 8-25, Problem 21_ (a) What is the combined kinetic energy of thc two blocks when the hanging block has fallen 0.120 m? (b) What is the kinetic energy of the hanging block when it has fallen that 0.120 m? (c) What maximum distance does the hanging block fall before momen- tarily stopping?



Answers

Two blocks, of masses $M=2.0 \mathrm{kg}$ and $2 M,$ are connected to
a spring of spring constant $k=200 \mathrm{N} / \mathrm{m}$ that has one end fixed, as shown in Fig. $8-69$ . The horizontal surface and the pulley are frictionless, and the pulley has negligible mass. The blocks are released from rest with the spring relaxed. (a) What is the combined kinetic energy of the cwo blocks when the hanging block has fallen 0.090 $\mathrm{m} ?$ (b) What is the kinetic energy of the hanging block when it has fallen that 0.090 $\mathrm{m} ?(\mathrm{c})$ What maximum distance does the hanging block fall before momentarily stopping?

In this problem on the topic of conservation of energy, we have two blocks of masses, two and four kg, that are connected to a spring, which has a spring constant of 200 newtons per meter, and one end of the spring is fixed. As we can see in the figure, the horizontal surface as well as the pulley are frictionless and the police has negligible mass. So the blocks are released from rest and we want to know the combined kinetic energy of the two blocks. When the hanging block has fallen 0.9 m, the kinetic energy of the hanging block after it has fallen through this distance and the maximum distance that the hanging block will fall before momentarily stopping. Now the initial height of the two M block has shown, is the Y is equal to zero level. In our calculations of the gravitational potential energy as that block drops, the spring stretches accordingly. Also, the kinetic energy is evaluated for the system and that is for a total mass moving, which has a total mass of three M. So the conservation of energy tells us that the initial kinetic energy plus the initial gravitational potential energy, is equal to the kinetic energy of the system. That's the gravitational potential energy of the system. So initially there is no kinetic and gravitational potential energy since the system is addressed. And this is able to be final kinetic energy of the system. That's the final gravitational potential energy, which is to m times G times minus 0.0 nine m plus elastic potential energy due to the stretching of the spring of a half K time zero point 09 m squared. If we substitute the mass of two kg, we get the kinetic energy of the system after the hanging block has dropped 0.9 m. Okay, to be 2.7 jules. For part B. We want to find the kinetic energy of the hanging block after this fallen through this distance. And the kinetic energy at the two M block represents a fraction of the total kinetic energy. So the kinetic energy of the two M block over the total kinetic energy for the system is equal to to M. V. Squared over two, divided by three a.m. V. Squared over two. And so this ratio of kinetic energies to over three. So we can see that the kinetic energy for the two M block is two thirds the total kinetic energy of 2.7 jewels, which gives us the kinetic energy to be one 0.8 jewels for part C. We want to find the maximum distance that the hanging block will fall before momentarily stopping. So we let why able to minus D. And solve A. D. And so again from the conservation of energy. Okay? I bless you. I is equal to Okay sis plus you for the system, so zero plus zero is equal to zero plus to M times G times minus D plus a half k D squared. If we use the mass M two B two K G, we can rearrange and solve for D. And we get this maximum distance to be zero 0.39 m.

In this question. We have a block of mess and, uh, 0.5 kg. She's on, uh, particular spring been constant. 80 newtons per meter. Okay. Actually, passing the question because we need to calculate the elastic potential energy. So, in the spring, uh, initially. And then after that, we drop a block from formulas height. And then I went to find a Mexico elastic potential energy start in the spring during the motion and and find a maximum distance. The first block moved down. Okay, so, uh, okay, do this part by part. So in Friday, first thing we need to find is that uh huh. It's a compression that Ah, the spring experiences. I think so. Like this. Why? Okay, so, um, from the free body diagram. What of the block? What? So we have mg, and then we have the spring force k x k y pointing up. Okay, so, um, the longest at rest. Okay, so m g you guys to ky and why he goes to energy. Okay. Yeah, I'm going to call this y one. Yeah, because there's a wine tour in Patsy, okay. And then, uh, elastic potential energy. All right. No, Yes, you got to have. Okay. Why? Once where and then, Uh, kay mg Okay, square. So you can, uh, and square G square over two K. Even you simplify, it becomes like this. And then you put in the numbers, okay? Calculated. You get 0.150 Jews. Okay, so this is the answer for part a. Then you go to part B. Paper B. You have a blog that's dropped, uh, from formulas. Height. Okay, that's where we draw the situation. Okay. And then it's going to drop 4 m height. Then after that, uh, to block combination, they stick together, and then the two block combination will go down. Okay? Compressors bringing for the case. So we want to find our maximum, Uh, potential elastic potential energy starts in the spring in this motion. Okay, so first, we need to calculate the speed before the blog collide. Okay, So, using conservation of energy for the falling block. Okay, so we have, uh, k i u I He goes to k f u f. Okay. The system is, uh, lock plus of a building block. So we have zero class, uh, m g h is equal to half MB square. Last zero. Okay, so the zero line in disk in this falling block type of system is where the, uh that where, though. Ah, Long is initially where the first bomb was initially arrested. So we can calculate the V to be, uh, square to G H B. Two times 9.8 times four square roots. Calculate this. You get it. $0.85 for a second, and then the two blocks collide and stick together. So in this case, momentum is conserved. Yeah. Oh, okay. So, using conservation of momentum here, we have a perfectly inelastic collision If you have em. This is N B. Is it good to to m b f D. So if you can see the ends you have, we have you guys to have weak. Yeah, which is 4.4 t. You know, it's for a second. Okay, Then we can calculate the kinetic energy, uh, after blocks, deli. Uh huh. Collision. He said this is equal to have to end B f square. Okay, So you completely if you get, uh, so you can consider to okay, 0.5 times 4.43 swear you get 9.8 Jews. Okay. And to find, uh, to find a maximum elastic potential energy, we're going to use conservation of energy. So what happens is that, um okay, so we have two blocks. They're moving down at 4.43 m per second and then Okay, so, uh, it's moving down. And after that, they are going to compress further. Okay, So just to draw better spring is longer, okay? And then they are. The spring is going to compress. Mm. But okay. It is initially compressed by, uh, by some distance, Okay. Just now, I didn't calculate the y one. Okay. And then it's going to go down further. Why two See? So just now, the y one p uh, why one is, uh, mg divided by K 0.5 times 9.8. Divide by 80. It is 0.613 Okay, if you are going to use this number later because we are going to use conservation of energy to solve for the for the total y Okay. Okay. So the total compression of the spring is equal to I'm going to call it. Why, And this is Why one? That's my two. Okay, then, uh, I'm going to use, uh, conservation of energy. Yeah, And then the system is, um is uh huh blocks? Uh huh. And spring. Mhm, then. Okay. So this is what I am going to, uh, calculate. Yeah. Oh. Okay. So, uh, do I plus a I go through, you ask. Thus chaos. Okay? To the eye. Okay. So that you I that I'm going to consider will be just 0.150 Jews. Okay. And then, uh, us ai is 9.8. That's why we calculate just now and then you f is. Uh huh. Here, wide square. And then, uh um, going to calculate, uh, that there's also a change in gravitational potential energy. Okay, so I'm going to call this, uh, to em the way, okay. And then, uh, zero. Okay. So because of the way we define our wire here, so we actually need to consider the, uh yeah, the m g y the negative. M g Y one from one block. Okay. Yes. Now. Okay. All right. So, uh, I'm just going to so n g y one k just put aside, uh, n g y one. Is it good to your point? Five times 9.8 times their point 061 tree. Hey, and this is actually 0.300 draws. Okay, So, uh, can rearrange the equation by substituting the numbers. So we have 40 White Square K is 80 80. Divide by two. You get 40 and then a white square. So we have an equation in quadratic equation in, uh, y okay. And then minus 9.8 km is 0.5 kg. Okay, so to m will be one kg g is like on a date and times y and then the other side. If you calculate this together, you get minus, um, nine point and 65 Okay, because 20 So why is he go to 9.8 plus minus 9.8 square? Uh, US four times 40 9.65 Okay, so it's minus 40 c, but she is minus negative. 9.65 So I put up last year, okay. And then if I buy, uh, 80. Okay, so calculate this. Okay, Then you get why to be 0.6 to 8 m or it across the other, and so it's going to be negative. But I still have to calculate this for the completeness of the step. Yeah, but this is not applicable. Yeah, we can calculate our healthcare. White Square. Okay. The maximum, uh, last date. Potential energy. Okay. Yes. Of ky Square? Yeah. Half times. 80 times 0.6 to 8 square when you get 15.8 juice. Yeah. Uh huh. Yeah. All right. So, uh, is this the answer for part B? Right. Okay, then in part C, you want to find a distance traveled by the first block? Okay. His own party. You want to find a distance? The maximum distance traveled by the first block. Oh. Okay. So this is actually why, minus y one, or actually just y two. So that's what, uh, the diagram suggests. So why is 0.6 to 8 and why one is your point? There were 61 tree to me. Um, do this 6 to 8, minus 0.613 Okay. And then you get your 0.567 m. Okay, so this is the answer for Patsy, and that's all for this question.

In this question. M one is M m toe is three m you want is You know you two is us zero V one. You we have to find out. Vito is two meter per second. Now, according to law, off conditionally, you know, went, um initial momentum musical to find no mentor. Judo is equal to em when we want plus m two v two. So everyone is minus m toe veto divide by m. One. When we get cleared. This minus 3 a.m. in tow to divided by am that is minus six meter per second. No. And this is the A part in which we have to find the velocity. In the be part. Your tell elastic. Put a shell energy equal Tok one plus care to half. Mm. When we won scare plus half m two, we do skill. This is half in tow. Jiro 0.35 in two minus six killed. Plus half in tow. Three in tow, 0.35 And you'll so scared. This comes 8.4 jewels in the sea part duh. Or is no energy is stored in the spring now in the deep park. The spring has, huh? Some displacement in the e part? Yes, it is. Gun soaked in the F part? Yes. Large forces are acting on in the deep part. The answer is yes.

We can say for party the maximum displacement from equal a gram. All of the energy is in the form of elastic potential energy. So here the total amount of energy would be equaling 1/2 times the spring constant k times the stretch with compression Max squared. And here Ah, we can say that the displacement from equilibrium is 2.40 meters. So to solve for our spring constant K, this would be equaling two times the total amount of energy divided by the maximum stretch. For the compression squared, this would be two multiplied by 47.0. Jules, divided by 0.240 meters quantity squared. And this is giving us brother squared. And this is giving us K equaling 1630 Newtons per meter, approximately rounding to three significant figures for this appearance or four card, eh for part B. Then at the equilibrium position, the spring is momentarily in at the spring is momentarily in. It's relaxed position, so it's not stretched or compressed. Uh, here we know that the elastic potential energy here we can say at equilibrium position Ah, where X is equal ing zero we can say that the elastic potential energy, potential energy associated with the spring equal zero. And so all of it is in the form of kinetic energy. So we can say that the kinetic energy at X equaling zero is going to be equaling 1/2 and the sub max squared, equaling the total amount of energy. And this is 47.0 jewels. So that should be our answer for part B for part C. Then we can say that if at the equilibrium possession V is equaling V max and this is equaling 3.45 meters per second than to solve for a mass of the block, this would be equaling two times the total amount of energy divided by the maximum velocity quantity squared. So this would be too multiplied by 47.0. Jules, divided by 3.45 meters per second quantity squared, and this is equaling 7.90 kilograms. So this would be our mass of the block for part C for a part D, Then we could say at any position the total amount of energy is constant. And so we can say that the energy is equaling the kinetic energy plus the potential energy associate with the spring or the elastic potential energy. This is equaling 1/2 M b squared plus 1/2 okay, x squared, where here x is equaling point 160 meters. So we're trying to find the velocity and we can see that then the velocity would be equaling the square root of two times the total amount of energy minus K X squared, divided by the mass. And so we can say for party the velocity would be the square root of two multiplied by 47.0 Jules minus 1000 630 Newtons per meter multiplied by 0.160 meters quantity squared and then this would be divided by the mass of 7.90 kilograms. And we find that the velocity at this point is 2.57 meters per second. So I'd be a velocity for part D at X equals 0.160 meters and then for part E again axes equaling 0.160 meters. However, here the velocity is 2.57 meters per second. And so the kinetic energy is gonna be again 1/2 M V squared. And so the kinetic energy is equal in 1/2 times 7.90 kilograms multiplied by 2.57 meters per second. Quantity squared, equaling 26.1. Jules, this would be our final answer for part E for part two f then Ah, here we have the elastic potential. Energy is equaling the total amount of energy eat minus the kinetic energy. Again here, X is equal in 0.160 meters. So this would be equaling 47.0 Jules minus 26.1 Jules. And this is equaling 20.9 jewels for the last six potential energy at X equaling 0.160 meters and then for park gi, the first turning point from rest at X equal and positive 0.240 meters. Here all of the remaining energy is in the form of elastic potential energy. So here we have 1/2 okay x squared, and this would be equaling the amount of energy minus the energy lost for that one cycle. And so this is equaling 47.0 Jules minus 14.0 Jules, giving us 33.0 jewels left essentially, and this is all potential energy, elastic potential energy. And so we can then say that the exposition now would be equaling the negative square root of two multiplied by 33.0. Jules, divided by the spring constant of 1630 Newtons per meter, extend the square root and this is giving us negative 0.201 meters. So, of course, a little bit less than 0.240 meters. But that is expected because there's some associate ID energy loss. So this would be our answer for part G. That is the end of the solution. Thank you for watching.


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