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Frictionless rope attached to the Iass runs Over 06 rest on horizontal surface: initially at rest, The HLei of m1 3kg is The system ottached hanging InS vTound: The...

Question

Frictionless rope attached to the Iass runs Over 06 rest on horizontal surface: initially at rest, The HLei of m1 3kg is The system ottached hanging InS vTound: The cocflicient and masslss pulley and descend Ad ultimutcly hits the Moin MId W sLuutn ground heard Iasses uU (hen allowed The sound of mz hitting the Juc the horizontal surface is 0.2. of kinetic friction bctwecn AIId the #OUld reach the ohscrver OVC down t0 the MOid 31 meters The time takcn lor m In /* uIL spe of sound as 340mn/8. tu

frictionless rope attached to the Iass runs Over 06 rest on horizontal surface: initially at rest, The HLei of m1 3kg is The system ottached hanging InS vTound: The cocflicient and masslss pulley and descend Ad ultimutcly hits the Moin MId W sLuutn ground heard Iasses uU (hen allowed The sound of mz hitting the Juc the horizontal surface is 0.2. of kinetic friction bctwecn AIId the #OUld reach the ohscrver OVC down t0 the MOid 31 meters The time takcn lor m In /* uIL spe of sound as 340mn/8. tu ground? (Asunc E(ond. How hlgh WILA Ine nbove ground level ) mna OLo axsutned that the: observer



Answers

(III) Two masses, $m_{1}=18.0 \mathrm{kg}$ and $m_{2}=26.5 \mathrm{kg}$ , are connected by a rope that hangs over a pulley (as in Fig. $8-47$ ). The pulley is a uniform cylinder of radius 0.260 $\mathrm{m}$ and mass 7.50 $\mathrm{kg}$ . Initially, $m_{1}$ is on the ground and $m_{2}$ rests 3.00 $\mathrm{m}$ above the ground. If the system is now released, use conservation of energy to determine
the speed of $m_{2}$ just before it strikes the ground. Assume the pulley is frictionless.

So the angular speed of the police is related to the speed of the masses. We can say that Omega would be cooing V over r and so we can apply the conservation of energy Energy initial equals energy final. And we can then say that we know that all objects have an initial speed of zero. So we can say 1/2 amps of a V initial square. This would be the initial translational kinetic energy of Mass A plus the initial translational kinetic energy of mass be plus the initial rotational energy plus the initial gravitational potential energy plus the initial gravitational potential Energy of mass sub B Rather we can say ASAP to I and then here you can say why sub one hi. And this would be equaling essentially the exact same thing but final. So it would be 1/2 im someday The final squared plus 1/2 I'm so happy. The final squared plus 1/2 Iomega final squared plus M c A G. Why someone final? Plus I'm sub b g. Why some to final? And so we essentially find that mass. Have you thought over here Massa b g h would be equaling 1/2 m sub a. The final squared plus 1/2 Massa be the final squared plus 1/2 times 1/2 em r squared multiplied by the final over our quantity squared plus m sub a g h. And so we can say that the final would be equal to the square root of two times m sub B minus m sub a times G h. This would be divided by M sub a plus m sabi, plus 1/2 times the mass of the pulley. And so we can then solve the final velocity would be equaling the square root. I have monitor two times 38.0 minus 35.0 kilograms multiplied by 9.80 meters per second squared multiplied by 2.5 meters. And then this would be divided by essentially a 38.0 plus 35.0 plus 1/2 times 3.1. The units are of course kilograms. This is equaling 1.4 meters per second. This would be our final answer for the final velocity. That is the end of the solution. Thank you for

Yeah. In this case, we have, uh, what's called an Atwood's machine. I'm not sure exactly why. Um, it's gets Mr Atwood. I assume, um, or Mrs that word gets their name attached to such a simple contraption. But that's what it's called. So it's basically just two masses hanging over a polio. Right? Um and so you have. We're assuming that this Ah, the cable connecting the masses is in extensible. So the distance that this one falls down must be the same distance that that one goes up. So we're asked to calculate what the velocity of this mass is. Um, just before it hits the ground and this mass is greater than this mass. So it's going to this one is going to fall on. This one is going to rise. We know that, actually, the velocity the velocity of have basically the velocity vector is has the same magnitude but a different sign, right? Because this is in extensible so and again. So that means that the distance this one fell must be the same as the distance. This one rose. Now I'm gonna set, um, gravitational potential to be zero right here in this initial state a, um, and they released from rest. So the total energy in the system is then zero. And that means the total energy in the system here is zero. So whatever, whatever kinetic energy was gained must have been lost in potential energy. And like I said, we know that the velocity of each of these bodies has to be the same. The magnitude of the velocity anyway, on. So that means that that's all we need for the kinetic energy is the magnitude or the speed. So we know then that, um, the energy in this state is zero. And that's the kinetic energy, plus the potential energy. Um, eso we have We can write down the kinetic energy of each particle and then the changing potential energy of each particle, and that must be equal to zero. So then we can use this relationship here, and so we get one. We get an equation where we can solve for VP in terms of the other parameters. So in the end, we get square root of two G h m one m two minus m one all over m one plus m two, and they give us some numbers here, Um, and if we plug in our values Ah, we get, um, that the values that we were given, we get the velocity just before this thing hits. The ground is 1.1 m per second. But because we got this in symbolic form here, we can actually do a little of what I call sanity checks to see if this answer makes sense. So one sanity check would say, Okay, one of them one equals m two. Well, that means the velocity zero, because basically, they're just an equilibrium here. If they're masses air equal, then neither either one is gonna fall and neither one is going to rise. So that makes sense. That's a good check on that. Ah, we can see if, um, one is zero that goes to zero. That goes to zero. These cancel out, which makes sense because we just have something under freefall and it's mass doesn't make, um doesn't change how fast it is falling in the gravitational field. So that makes sense. And we get the normal normal relationship that we just have. The velocity of a particle in a free fall is square to GH or square to G H where h is the distance that it's fallen. So that makes sense. One other weird kind of one that maybe doesn't make took a lot of sense to say. I'm too. Is zero. Well, then then we basically get a square root here. Right? We get a new imaginary number minus square root of minus two GH. Well, that's because we basically assumed that this one was moving up way We made that assumption here, and so we needed we would need to actually, um, we made that assumption essentially here. Um, you know, here that this one was moving up, that h is positive. Um, in fact, eight would be negative. Um, in that case, and then we would have a real number in there because we had, um, a negative age times minus one. So again, we essentially assumed that this one was going up. And that's why we would if h we're positive than you know, we get a, um, nonsense of quite answer. But that means that basically, we're just saying that things don't fall upwards. Um, h would have to be negative of this mass. Where? Zero. Okay,

Hi. In a given problem, there are two blocks, one of which has been put at the people talk. And it is having Amoss m one as 8100 g. A never block is suspended with the help of trade which is passing over our frictionless fully. And it's mosque iss m two which is being given as 6.0 candy. It's where it will be acting vertically. Don't work creating attention t in the strength in the trip Vato this look yes, I m one g So the normal reaction exerted by the tabletop on the block is in which is also equal to and one G as the block will be moving towards right. So force of friction with the acting Do you live given by mule in tow end here Mu is missing as the blocks are moving like this. The suspended block is moving downward with an ex solution is so This lock will also be moving towards right in the same exploration it. So if we use free body diagram off this second block having mass m toe, we get to know MDG minus Steep is the nerd for 700 and using Newton's second law of Motion. It is put people to I'm going to it. So tension here comes out to be m G minus. I m do a where you can see becomes Anto G minus t. But before that we will have to find the acceleration in the system. Of course, who find this acceleration? We will use kinda Matics because you should speak has been given to us as 0.900 meet up for a second and the system is finally coming to rest. So finally, speed is zero and the distance moved by the system by the hanging block and the block on the table. It's same which is giving us 2.0 meter. So used her equation off. Aromatics, which saves the F A square is equal to the I swear minus school A s, so it is converted into zero is equal to 0.9 for the whole slab minus to pay into. So acceleration here comes out Toby 0.81 divided by four What we can see. Acceleration is 0.20 meters per second squared. So now going back. So this equation we find Engine T in the block in the change in the church as M two into G minus eight. So this is M two G minus C. We know the mass of the block as six G 900 minus 0.2. So the tension comes out of the 57.6 million. Now we used pretty body diagram off the block. Put on the table talk. It gives us t minus four. So friction as I mean 2 a.m. one into a and embodies eight. So it becomes 80 or you can say it into 0.2. So tension, you know, 57 564 Force of friction, it becomes Muto. End an N s A G means it into 9.8 is going to 1.6. So new into a into 9.8 comes out Toby 57.6 minus 1.6, which becomes 56. So the coefficient of kinetic friction between the tabletop and the block over it becomes, if they're six divided by it in 29.8 and here comes out to be 0.7 months. Since this is answer for the given problem. Thank you

So I think very beneficial to draw this out. So we have a pulley with radius are and we have one mass here. This is going to be massive, eh? And then we have one mass here. This will be massive bee and we have this height difference here. This is going to be considered. Um h So we have our givens mass of a is going to be equal to thirty two kilograms mass of is going to equal to thirty six point zero kilograms. Rather rather not thirty six, thirty eight and we have the radius is equal to point three one one meters. We have the mass of the pulley equaling three point one kilograms. And then we have height of two point five meat of two point five meters. So we should first find out exactly what what our energy transfer is here. So let's first we have the the moment of inertia of this pulley. So this pulley is going to be a mom. He r squared over two. So I'll be the moment of inertia of the pulley. And then we know that energy initial equals energy final. It's about all we know, right? now so we can right a relationship and say that gravitational potential energy of mass B equals thie kinetic energy, rotational of the pulley plus kinetic energy, translational of a plus kinetic energy, translational of beer plus gravity potential. The gravitational potential energy of a at the end. So let's get a new work because this is a bit long and we have the mass of the massive B times G h equals, huh? The moment of inertia of the pulley times the angular velocity of the pulley squared plus one over to mass of a velocity squared plus one over to mass of B velocity squared, plus one over two. Rather rather, this right here would be the gravitational potential energy of a. So this would be eh G h So at this point, we need to solve. And we can weather plug in everything that we know. We have thie gh, and we can really only effect to this term. So one over two and then this is just going to be one over two massive the pulley radius squared. And then this will be V over r squared plus one over to mass of a V squared plus one over to mass of B the squared plus mass of a gh. At this point, we can salt weaken, simplify. And we know that this term right here is simply going to be equal to one over four mass of the pulley v squared because this are is going to cancel out this are this too will cancel out whether this too this too will cancel out. This too will just be simply multiplied by the one over to factor here. So again, this would be one over four times the mass of the pulley v squared. And at that point, if we substitute this and for this term right here, we can solve for V because we will simply equal to gh mass of B minus mass of a all divided by massive a plus mass of B plus. Ah, massive the pulley over too. And this is going to be to the one half power and we can substitute so too nine point eight two point five on this will be thirty eight minus thirty two, all divided by thirty two plus thirty eight plus three point one over two, all to the one half power and we find that V is going to be equal to two point zero two seven one meters per second or we can say approximately two point zero three meters per second. So this will be your answer for the final velocity of these system after it has. After the initial conditions are released, we simply have to realize I think the main thing is just do not forget that there's gravitational potential energy from mass of A because it's very easy to just say that gravitational potential energy equals kinetic energy. However, we need to account for the fact that there's some gravitational potential energy in Block A after after the after Block B is released. So just keep that in mind. Let's keep this term in mind and again that final answer to point zero three meters per second. That's the end of the solution. Thank you for watching


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