Two blocks with masses mA and mB are connected by a masslessstrin...

Question

Two blocks with masses mA and mB are connected by a masslessstring over a massless, frictionless pulley. Block B, which ismore massive than block A, is released from height h and falls.a. Write an expression for the speed of the blocks just as blockB reaches the ground.

Accepted Answer

Hey, everyone. So this is a conservation of energy problem with a pulley. Let's see what they're asking us. A five kg bucket of water rests on a horizontal surface tied to a vertical rope. The rope of negligible mass passes through a frictionless pulley. The other end of the rope is attached to a six kg weight. The weight is dropped from a height of 2.5 m. Find the speed of the weight just before it hits the floor. Our multiple choice answers here are a 1.5 m per second. B 2.1 m per second. C seven m per second or D 23. m per second. OK. So let's start by drawing a little diagram of what's going on here. So we have a pulley, we have our bucket of water and then we have on the other side, the weight And we know that that weight is a distance of 2.5 m above the ground. We can recall that our conservation of energy equation is given by U I I see it's very U I plus K I equals U F plus K F. In turn, we can recall that you are um potential energy is given as M G H and K, our kinetic energy is given as one half M V squared. So when we write out the total equation for this problem, we have to keep in mind that this entire system, so the bucket and the weight together, that is the system um where we have that conservation of energy. So we need to look at the initial kinetic and potential energy of both the bucket and the weight. And then again, the final potential and kinetic energy of both the bucket and the weight. So it's going to look like a lot when we write it out, but we will work through it and uh show how things can be simplified. So let's take our initial potential energy, we have M B multiplied by G multiplied by H B I plus M W G H W I plus one half M B V B I squared plus one half M W V W I squared. So that is our initial side of the equation. And then we'll do the same for our final side. So we have M B G H B F plus M W G H W F plus I'll do our kinetic energy one half M B V B F squared plus one half M W V W F squared. OK. So there is a lot going on here. Let's take each term one by one, figure out what we were given in the problem what we need to solve for and do any kind of simplification that we can do. OK. So one by one M B, the mass of the bucket was given to us in the problem five kg G gravitational constant, we can recall 9.8 m per second squared H B I. So the height, the initial height of the bucket that's going to be zero because it starts resting on a horizontal surface. I equals zero m. OK. The mass of the weight We were given and the problem as six kg gravity we already defined and the initial height of the weight We were given in the problem as 2.5 m. We're off to a good start looking at the kinetic energy terms, we already defined the mass of the bucket. So the initial speed of the bucket and the initial speed of the weight are both going to be zero because we start before our initial condition is before the weight is dropped. So the bucket and the weight are at rest. So V B I equals V W I equals zero m per second. Ok. Let's look at our final conditions here. You've got H B F Is going to equal 2.5 m because the weight and the bucket are attached by a string. If the weight moves 2.5 m, the bucket, the if the weight moves 2.5 m to reach the ground, the bucket must also move 2.5 m. So that's how we know that HBF equals 2.5 m. H WF is zero m because we are looking at it, um looking at the speed right before it hits the floor. So essentially 0m. And then V B F and V W F, we're asked to find the speed of the weight. So that's V W F But again, because they are connected, the bucket and the weight are connected by this rope, they're going to move the same speed. So V B F equals V W F and we can just call that V F and that is what we are solving for. So we've worked through this pretty complicated equation and we can see now we have every value we need to solve it to solve for V F. First. Let's simplify, let's look at our zero terms and simplify those out. So H B I is zero, which means this whole term goes to zero. The B B I V W I are both zero. So these terms go to zero and then H W F is zero. So that also goes to zero. So we can rewrite this equation. We have M W G H W I equals M B G H B F plus one half M B V F squared plus one half M W V F squared. Again, recalling here that the V B F and B W F right were just those are the same. So we are just calling those V F. OK. So from here, you can choose to continue um kind of manipulating the equation around until you have all of your variables with V F isolated. I think it's a little bit easier. We have all of the values to plug in and I think it'll be a little bit easier to work with actual numbers at this point. So I am going to plug in um the values that we have and solve for the terms except for you know, V F. So we have six kg multiplied by 9.8 m per second squared. multiplied by 2.5 m equals five kg multiplied by 9.8 m/s squared. Multiplied by 2.5 m plus one half, multiplied by five kg, multiplied by V F squared plus one half, multiplied by six kg multiplied by V F squared. So we plug that in and we get 147 jewels Equals 122.5 Jews plus 5.5 kg multiplied by V F squared. And so this is a little bit easier now to isolate V F. So VF is going to be equal to the square root of 147 joules -122. joules all divided by 5. kg. We plug that into our calculator and we get 2.1 m/s. And that is the answer to this problem. Let's look at our multiple choice answers here that aligns with answer choice B so that's all we have for this one. We'll see you in the next video.

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Two blocks with masses mA and mB are connected by a masslessstring over a massless, frictionless pulley. Block B, which ismore massive than block A, is released from height h and falls.a. Write an expression for the speed of the blocks just as blockB reaches the ground.

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