More Conservation of Energy Problems - Online Tutor, Practice Problems & Exam Prep

In a system with two blocks connected by a light rope over a pulley, the conservation of energy principle is applied to find the final speed of the blocks. The potential energy of the falling block is converted into kinetic energy, accounting for work done against friction. The equation derived is $\sqrt{\frac{4gh(m2-m1)}{2m1+2m2+m3}}$. The final speed calculated is approximately 6.47 m/s.

Speed of blocks on a pulley (Atwood's Machine)

Hey, guys. So in this video, I'm going to show you how to solve this very popular question using conservation of energy. So this is a conservation with energy with rotation question. Let's check it out. So we have 2 blocks connected by a light string. Here are the 2 blocks. A light string means that the mass is 0 and the string is ran around the pulley as shown. So the string is like this, the red line. This setup, by the way, is called Atwood's machine just in case your professor mentions it. It's a very classic problem pulley with 2 objects hanging from it. The blocks have masses 35. So I'm going to put a 3 here, and I'm going to put a 5 here. The pulley is a solid cylinder. This is telling us the shape of the pulley so I can know the moment of inertia equation to use, which is going to be for a solid cylinder ix=12mr2. I'm actually going to call this m3 because I have 2 objects. m1,m2, and I'm going to call this 3. And then r2. This is the only object that has a radius. So I don't have to say r3. I just say r. The mass m3 is 4 and the radius is 8. Now if you wanted, you could already calculate I. We're going to do this a little bit later. Some interesting stuff happens if you leave everything in terms of letters as I will show you. It is free to rotate about a fixed perpendicular axis through its center. So what does that mean? So let me get a little disc here. The idea is that the pulley is a disc. It's a solid cylinder. There's an axis through the center and perpendicular. So through the center and perpendicular means making a 90-degree angle like this with the disc. It's free to rotate. So when you put the masses of different, when you put the objects of different masses, it's going to tilt towards the heavy one. The heavier one and but the axis is fixed. So the disc itself isn't going to move. Imagine that it's like attached to a wall or something and it doesn't it can only spin but not move. So that's what that means. The whole thing is released from rest. Initial velocity is 0. And m2 is at a height of 5 meters above the ground initially. So m2 is 5 kilograms but it also has an initial height h2initial=5. When you release because it's the heaviest one, it's the heavier one. It's going to go like this. Right? And it's going to hit the floor. So its final height is 0. Alright? The question is, what is the speed of m2 just before it hits the ground? So what is v2final? That's what part a is asking us. What is v2final? Okay. v2initial=0 because the system starts from rest. The second question part b asks, what is the pulley's speed? Now when we talk about the speed of the pulley, we're talking about omega. The pulley doesn't have a v. It just rolls around itself. So what is omegafinal? Again, it's the only object that rotates so we don't have to say omega3final. We could just say omegafinal. Okay. Before we start, we're going to use conservation of energy, by the way. But before we start, I want to point out that this is a system. Objects are connected and connected objects always move together. They have the same velocity. So v1 is at all times the same as v2 of the 2 blocks. Sometimes you you you see this called the velocity of the system, vsystem. Right? Or simply v because there's no point in differentiating 1 and 2. So we just call them both v. Also, whenever a rope is connected and pulling, on a pulley, we can write the equation. So this is point number 1. Point number 2 is that we can write the equation, vrope=romegadisc. Whenever rope pulls on the disc, where r is the distance to the axis. It's the distance between where you pull on the rope, where you pull on the rope and the axis of rotation. In this case, the rope pulls on the disc from both sides at a distance of the radius. Right? So all the way at the end means that the distance is the radius. Okay. So I'm going to be able to write that simply v=bigrOmega. Okay. So I have these 2 extra equations to use. And by the way, if v=romega, I can just add that right here. So it's really just one big equation. Okay. You can think of this as all the velocities the same. v=v,whichequalsromega. Cool. Alright. So let's go ahead and write our conservation of energy equation. Kinitial+uinitial+worknonconservative=kfinal+ufinal. What's going to make this question not harder but sort of longer and more annoying is the fact that I have 3 objects. I have to worry about the energy of all 3. So when I ask, is there a kinetic energy in the beginning, you have to think about all 3 objects. And actually in the beginning, the system is not moving so there is no kinetic energy. Potential energy is 3 of them. And for now, what I'm going to do is I'm going to write all 3. Uinitial1,uinitial2,uinitial3. And we'll talk about that in just a second. There's no work non-conservative because there's no work done by you. You're just watching. There's no work done by friction. Okay. And kinetic final, everything's moving just before we hit the ground. Right? This guy is going down. This guy is going up and the disc is spinning. So I have kinetic final for all 3 of them. And I'm going to write the potential file as well. Uf1,uf2,uf3. Now let's analyze this real quick. Does the first guy right here have potential in the beginning? The answer is no because it's on the floor. But it does have potential energy at the end because they flip. Right? The second one has potential energy at the beginning because it's up here, but doesn't have it at the end because they flip. Okay. So notice how 2 has it here and then one has it here. What about the disc? The height of the disc doesn't change. In the beginning, it's up here. At the end, it's up here. So we can cancel the potential energies like this. All 3 of them have kinetic energies, but it's not enough to know that it has kinetic energy. You have to know what type of kinetic energy it has. Well, the blocks are moving up and down. So this is linear motion. So they have linear kinetic energy, but the disc is spinning and it only spins around itself. It only has one type of motion. And so it has rotational kinetic energy. Cool. Now we end up with 5 terms. We're going to expand all of them. So this is going to be mgh and it's for object 2. So I'm going to put a 2 initial. It's the only energy we have in the beginning. Here we're going to have this is linear so it's 12mv2+12mv2+12Iomega2. Let's put our coefficients here. This is the first mass and it's final. First second mass and it's final. This is the moment of inertia, the 3rd mass. I don't really have to put a 3 there, because it's the only thing that has a moment of inertia. And then we have the gravitational potential energy, which the only one we have is mghfinal. This is for the first mass. Cool. Now if you look through this, you might be wondering, can I cancel some stuff? You actually can't cancel anything. You have m's everywhere, but you don't have one here. And more importantly, all the m's are different. Right? So you're not going to be able to cancel the masses because they're all different. Let me clean that up. And what I want to do here is I want to sort of derive an equation. So I'm not going to plug in numbers until the end because I want to show how some stuff cancels. Okay. So it's going to look really nasty, but, you know, if you're doing this on a test and your professor doesn't mind, you could start plugging numbers already. Alright? But just check this out for, this one time and see some of the things that are going to happen here. So one of the things you want to do going forward is you always want to replace I with the I equation, which is right here. And you always want to replace omega with v. Remember we talked about this? Whenever you have a question that has a v and an omega, you're always going to you always want to replace rewrite your omega in terms of v. And that's so that instead of having v and omega, you have v and v And that's better because it's fewer variables. The way you do that is you get this equation right here and you say, omega equals v/r. So that's what you do. Okay. So I'm going to do that. There are 2 things to do here. 2 things to expand. Nothing else can be expanded. So you're just going to leave it alone. So I'm going to rewrite this whole thing and then expand this when I get here. Okay. m2gh2i. All I'm doing now is rewriting this. It's kind of annoying. Alright. Stop right there. Let's make sure I'm going to leave some space. Finish writing this. Okay. That was just rewriting. Now I have to actually, slow down a little bit here. For I, I'm going to plug 12m3r2. And here I'm going to plug, for Omega, I'm going to plug v/r. So v/r and it's big r. Notice what happens, the r's cancel, but really nothing else is going to cancel. Right? What I like to do at this point is because I don't like fractions, I'm going to multiply by the smallest number that will make this, make all the fractions go away. Basically, you have a half here. You have a half here. Here you have a one over 4. If you multiply that, that's the lowest denominator. So I'm going to multiply this by 4. Okay. And that way I end up with a 4 here. 4m2gh2y. This becomes a 22monevonefinalsquared. By the way, one more thing before I continue. All these these are the same. Remember we talked about that. So there's really no point in writing v1final. It's just vfinal. Okay. Plus 4 multiplies here. This becomes a 22m2vfinal. 4 multiplies against the 1 quarter, so this becomes a 1. It's going to be m3vfinalsquared+4moneghonefinal. This is for the first mass right there. Alright. So remember what we're looking for. Okay. This is extra painful because we're not pulling the numbers just yet. We're looking for vfinal. Notice that we have vfinals everywhere and no omegas. That's what she wants. Okay. You want to have a bunch of vf's everywhere and no omegas. Notice also if you do sort of an inventory of everything you have, you know the masses, You obviously know g and you know the heights. Right? The initial heights of the second block is 5 and the final heights of the first block is 5 as well. Right? Because this guy lost 5 of heights. So this guy could gain 5 of heights. So I'm going to put it here that both of these numbers are 5. You know everything. So at this point you can plug in numbers and solve. But I'm going to solve this with the letters instead. And what you would do is you combine all the v's. So there's vf's squared everywhere and I can combine the masses. 2m1+2m

Blocks on a rough table and a pulley

Hey guys, I hope you tried out this practice problem. Let's check it out. So here we have 2 blocks connected by a light rope, which passes through a pulley as shown. So here's the rope right here. It is a light rope, which means the mass of the rope is negligible. Okay. The pulley is a solid cylinder. So the moment of inertia of the pulley will be half m₃r². Pulleys are always solid cylinders. I give you the mass. Mass, I'm going to call this m₃. Okay. Let's call this guy m₁, m₂. Let's call the pulley m₃. M₃ equals 10 and the radius is 2. The 4 kilogram block is on a horizontal surface. It shows here. And the surface block coefficient of friction is 0.5. So there's friction here, μ equals 0.5. The system is released from rest. The initial velocity is 0. And the 6 kilogram block, with the 6 kilogram block initially 8 meters above the floor. So the initial height of m₂, h₂initial is 8. Okay. And I want to know what is the speed that the 6 kilogram block will have just before hitting the floor. So just before hitting the floor means that this guy will have a final height h₂final of 0. And I want to know what is v₂final. V₂initial is 0 because it starts from rest. Alright. So that's what we're looking for. Remember that this is a system. Everything moves together. Therefore, all the velocities are the same. Had I asked for the final velocity of the 4 kilogram, it would have been the same exact thing as the final velocity of the 6 kilogram, or had I asked for the final velocity of the system. So I'm going to write that v₁ equals v₂, which equals v_system, or I can just write this as v. Furthermore, that's point 1. Point number 2 is the rope is connected to the pulley at its edge. So I can write that v_rope equals Rω_pulley or disc. Okay. Now because it's connected at the edge, r equals big R. Little r is the distance between the center of the pulley and where the rope connects. The rope connects all the way at the edge of the pulley. So it's a distance of radius, which means this becomes big Rω. Okay. V equals big Rω, which means I can put it right here. So this whole thing here is that this big equation here that connects all these things. Now, obviously, this thing is going to move like this, this moves like this, and this moves like this. Because I have changing heights, I have friction. So I'm thinking the work done by friction, velocities are changing. I'm going to use the conservation of energy equation. So k_initial + u_initial + work non-conservative equals k_final + u_final. There's no kinetic energy at the beginning because these objects, the whole system is at rest. Let's look at potential energy. Now instead of writing all 3 of them, we're going to talk about it. So does this guy have potential energy in the beginning? It doesn't. Or even though it's above the floor, the 4 kilogram is moving sideways. The height doesn't change, which means whatever potential energy it has in the beginning, it's going to have at the end. The 2 of them would cancel in the equation. So you can basically ignore the potential energy of this guy. The same thing with the cylinder. The cylinder stays in place. It doesn't change potential energy. So the only object that has potential energy that's worth writing down is the 6. Okay. So I'm going to say that the only potential energy is going to be m₂gh₂initial. What about work non-conservative? Well, there's no work done by you. You're just watching. You release this thing from rest. But there is work done by friction because friction is acting right here. Okay? I'm going to say that there's the work done by kinetic friction on object 1. Alright? What about kinetic energy final at the, kinetic energy final? There is kinetic energy at the end. All 3 blocks or all 3 objects are moving in some way. The top one is moving horizontally. The 6 kilogram is moving vertically so they're both moving linearly. The disk spins, so the disk has rotational energy. So I'm going to write half the first block here has linear. So half m₁v_final². The second object right here has linear motion as well. So it's half m₂v_final². And the third one has, the disc has moment of inertia, so it's half Iω_final². There is no potential energy at the end. Remember, the block has no potential energy because it doesn't change height. The disc doesn't change height, so it has no potential energy. But this guy hits the floor at the end, so it has no potential energy. So I'm just going to put a 0 here for potential energy. One last thing, one quick adjustment here. These v's are really the same. So I'm just going to write v_final. Okay. And that's actually what we're looking for, v_final, v_final. There are 2 of them. So we're going to be able to combine them. But before we do that, we have to remember, we have to replace I and ω in here. So let's do that. I have m₂gh₂initial. I also have to expand the work done by friction. And I'm going to go off to the side here real quick, so we can do that. Work done by friction kinetic is negative friction kinetic times distance. I want to remind you that friction is μ normal. And in this particular problem, the 4 kilogram is on a table like this. So mg is down. Normal is up, normal equals mg. So I'm going to replace μnormal with μmg. And lastly, I'm going to put this f over here. So the work done by friction, the work done by friction is negative, μmg, that's f right there, d. So I'm going to get this whole thing and put it here. Negative μmgd. Alright. Let's keep going. So here I have half m₁v_final², half m₂v_final² plus half the moment of inertia is another half m₃r². And remember, we're replacing we're supposed to replace ω, right, from this equation here. We're supposed to replace ω with v over r. That's one of the most important parts of these questions. I need you to remember that you're going to replace ω with a v. Okay? So this becomes v over r. It's v_final since it was ω_final. Notice that this causes the r's to cancel. What this also does is now you have 3 v_final's and you're going to have to combine them. Okay? Here, you can't cancel masses. You're not able to cancel the masses because you have different masses everywhere. Okay? This one here, by the way, it's mass 1. I forgot to write it there. All right. So if you look around, you notice that you have all these numbers. You have the masses. Obviously, you have gravity. You have mass, you have gravity, you have mass, you have mass, you have mass. We're looking for the velocities. I'm giving you the coefficient of friction. So the two guys we haven't talked about here are the initial height that you drop or the total height you drop or what the initial height was and the distance that the block, the 4 kilogram block moves, and I hope you realize that they are the same, right? This block is moving down and pulling this guy, so they're both moving the same distance. So this distance here, d, is the same as this height. So what I'm going to do is I'm going to call this big H, and I'm going to call this big H. And it's the same exact number. Okay? So I'm going to have m₂gbigH - μm₁gbigH. And on the right side, what I'm going to do is combine all the v's. So notice that you have this here in front of the v's. This one's like this. Now I usually multiply this by a number to get rid of the fractions. I haven't done that yet, but I'm still going to do that. V_final². Then I have half m₁ + half m₂ + 1 quarter m₃. Okay. So what we're going to do here is multiply everything by 4 to get rid of these fractions. And here, I'm going to end up with I'm just going to do this here. I'm going to put a 4 in front, a 4 in front. These guys will cancel here. Don't put a 4 here. Right? Do not put a 4 there. We're going to already distribute the 4. So this becomes a 2, this becomes a 2, and this becomes a 1. I want to point out here if you want to clean this up, I got gh, I got 4gh and 4gh that's common, so I can factor it out. 4gh. I have m₂ minus μm₁. And then on the other side, I have v_final², 2m₁ + 2m₂ + m₃. Okay? I'm going to move the masses to the other side, take the square root, and we are done. The final will be the square root of 4gh(m₂ - μm₁) divided by (2m₁ + 2m₂ + m₃). And that's it. Okay. We're going to plug in the numbers in just a second. Now this is the hardest way to do this in terms of it's harder. What I mean by that is that it's harder to take it all the way to the end without plugging the numbers. I wanted to show you the harder version of how you might be asked this question, which is derive an expression. Right? Now if your professor doesn't ask you for an expression, if he asks you just to find a number, and if he doesn't care that you plug in numbers at any points, you probably want to plug in your numbers somewhere here. Okay? So don't worry about taking it all the way to the end unless you have to. Alright. So I'm showing you how to do that in case you do. But if you don't, don't be a hero. Start plugging in numbers here. It's much easier than carrying all these letters around for a long time. Cool. Now if you were to plug all of this, I have this already done. It's 6.47 meters per second is your answer. Cool? That's it for this one. Pretty common type of problem, so make sure you understand how to do this. Let me know if you have any questions.

Speed of a yo-yo

Hey, guys. So here's another classic rotation question that we're going to use conservation of energy to solve, and it's a yo-yo question. Right? So, we have a simple 100-gram yo-yo that we're going to release from rest. Now, "simple" just means that you're going to be able to make some assumptions to simplify the yo-yo. Yo-yos are actually more complicated than how we're going to solve them. But here we're going to simplify. That's what "simple" means. It just means, you know, go nuts with simplifications. Alright. So, mass equals 0.1 kilograms. It starts from rest, v_initial equals 0. It falls and rolls. Yo-yos do that. Right? So they're falling and rolling on the way down, unwinding the light string around its cylindrical shaft. So as it falls, it unwinds a light string. The yo-yo has a string around it. The "light" string means the mass of the string is negligible around its cylindrical shaft, and that's because a yo-yo has a yo-yo has a thing in the middle. The cable, the string is here, but then the yo-yo usually has sort of an outer casing like that. Right? The idea is that what matters is this inner radius, not the outer radius. The outer radius just covers the outside. So effectively, we're going to just worry about this and say that a yo-yo looks like this. Okay. Let's actually put the little string here. Now if the string is here and you release a yo-yo, it's going to fall and it's going to roll like this. So if the string is on this side, it's going to go like this. This is the velocity of the center of mass, and it's also going to spin with the omega. Okay. So, the radius of the inner, which is what matters, is 2 centimeters. So 0.02 meters. It says if the yo-yo can be modeled after a solid disc, in other words, treat this thing here as a solid disc, right, which is what we're going to do. In other words, I will be half m r squared because that's the I for a solid disc. I want to know what is its line

A small 10-kg object is connected to the right end of a thin rod of length 4 m and mass 5 kg. The rod is free to rotate about a fixed perpendicular axis on its left end, as shown below. The rod is initially held at rest, horizontally. When the rod is released, it falls, rotating about its axis, similar to a pendulum. What is the speed at the rod's center of mass when the rod is vertical? BONUS:What is object's speed when the rod is vertical?

• A. 4.4 m/s
• B. 4.9 m/s
• C. 5.6 m/s
• D. 6.1 m/s

Here’s what students ask on this topic:

How do you apply the conservation of energy principle to a system with two blocks and a pulley?

To apply the conservation of energy principle to a system with two blocks and a pulley, you need to account for the potential energy, kinetic energy, and work done by non-conservative forces like friction. The total mechanical energy of the system remains constant if only conservative forces are acting. The equation used is:

$K_{\mathrm{initial}}+U_{\mathrm{initial}}+W_{\mathrm{non-conservative}}=K_{\mathrm{final}}+U_{\mathrm{final}}$

Here, Kinitial and Kfinal are the initial and final kinetic energies, Uinitial and Ufinal are the initial and final potential energies, and Wnon-conservative is the work done by non-conservative forces.

What is the role of friction in conservation of energy problems involving pulleys?

In conservation of energy problems involving pulleys, friction plays a crucial role as a non-conservative force. It does work against the motion of the blocks, converting some of the mechanical energy into thermal energy, which is not recoverable. The work done by friction is calculated using:

$W{\mathrm{friction}}_{}=\; -\mu Nd$

where μ is the coefficient of friction, N is the normal force, and d is the distance over which friction acts. This work is subtracted from the total mechanical energy of the system.

How do you calculate the final speed of blocks in a pulley system using energy conservation?

To calculate the final speed of blocks in a pulley system using energy conservation, you set up the energy conservation equation:

$K_{\mathrm{initial}}+U_{\mathrm{initial}}+W_{\mathrm{non-conservative}}=K_{\mathrm{final}}+U_{\mathrm{final}}$

For a system with two blocks and a pulley, the equation becomes:

$4\mathrm{gh}\left(m_2-m_1\right)=2m_1+2m_2+m_3$

Solving for the final speed v_final:

$v$_final = 4gh(m2 - m1)2m1 + 2m2 + m3

Plug in the given values to find the final speed.

What is the significance of the moment of inertia in pulley problems?

The moment of inertia is significant in pulley problems because it accounts for the rotational inertia of the pulley. It affects the system's total kinetic energy. For a solid cylinder pulley, the moment of inertia I is given by:

$I=\frac{1}{2}m{R}^2$

where m is the mass of the pulley and R is its radius. The rotational kinetic energy of the pulley is:

$\frac{1}{2}I{\omega }^2$

where ω is the angular velocity. This term must be included in the total kinetic energy of the system.

How do you handle potential energy changes in a pulley system with blocks?

In a pulley system with blocks, potential energy changes are handled by considering the height changes of the blocks. The potential energy U is given by:

$U=mgh$

where m is the mass, g is the acceleration due to gravity, and h is the height. For a block moving vertically, the change in potential energy is:

$\mathrm{\Delta U}=mg\mathrm{\Delta h}$

where Δh is the change in height. Only the block that changes height contributes to the potential energy change in the system.