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$\bullet$At a construction site, a 22.0 $\mathrm{kg}$bucket of concrete is connected overa very light frictionless pulley to a375 $\mathrm{N}$ box on the roof of a building.(See Figure $5.49 . )$ There is no appre-ciable friction on the box, since it ison roller bearings. The box startsfrom rest. (a) Make free-body dia-grams of the bucket and the box.(b) Find the acceleration of thebucket. (c) How fast is the bucketmoving after it has fallen 1.50 m(assuming that the box has not yet reached the edge of the roof)?

a) $\mathrm{Box}$ Bucketb) 3.6 $\mathrm{m} / \mathrm{s}^{2}$c) 3.28 $\mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 5

Applications of Newton's Law

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

University of Michigan - Ann Arbor

Simon Fraser University

McMaster University

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Okay, so let's start with the free body diagrams for the box and the bucket. So first the box you have Ah, the weight of the box Toby Box backing down downwards of the tension of the poli acting horizontally. And you have a normal force exerted by surface of bush in boxes on the box, acting directly upwards. And your acceleration is the box of selfishness to the right. Then you have. And then for the bucket. Ah, you have cross the weight of the bucket backing down Winston called. Gonna shorten that too. Buck, I'll be back with a bucket and the intention and the of pulley from by the poet acting upwards and acceleration in this case is hacking downwards to pull the bucket down. Thie idea here is that this day and this air are the same. And this is because because they are the pulley that connect some war. Give these to objects the same acceleration given that there's no infection. Okay, so this is a free body diagrams and cells when we want what we want. The acceleration of thes objects. Ah, acceleration of the bucket will be same exception of boxes. He just noted. And so we used nothing. Second line, we resolved forces. Let us try tryingto extraction. Um And so for the from the box we have the tea is the only force in the ex direction, and that will be equal to mass of box times acceleration. Okay, Come into question one on then from the from the from the bucket we have well, from the bucket we d'oh! We was all forces in the white direction. His acceleration here's me in the white direction is I see nothing in the expression happening here. So we have t minus the weight of the bucket up you Buck he called her Massive The Bucket again Embark times acceleration. Okay. Ahh! And that two No. Okay. And so, um and so and so Ah, a convention we can define here is that no downwards is no positive. So Ah, this will actually be a negative t plus w buck or W Buck Dynasty called Sam back times, eh? Because acceleration is now we're them. That's the positive direction. And so what we can do here is we can use one and two. And so one just says that tea is equal to em box times. And that is equal to two. Which says that tea is equal to and buck times, eh? Plus w buck weight of the bucket. Okay, All right. And so what we get? No. Is that no embarks? Time's a rh is just em back times a plus way of buck. Okay? And so we can take a common that comes out to be w buck weight of the bucket over, uh, and box plus embark mass mass is off the box and the bucket combined and said w buck is just massive the bucket times acceleration due to gravity. And we have the same stuff here. Um and so let's let's let's just put that in. We have a massive bucket which has given us 22 kilograms. Multiply that by 9.8 meters per second squared Divide that by less of the box plus mass of the buckets of massive blood back in again 22 kilograms. The mass of the box is not given, but we are given the way and again weight equals and she so the mass must be the weight of the bucket over accession to the gravity. Then did 3 75 nipples over 9.8 meters per second squared. Wait and what we get is that acceleration is about 3.58 meters per second squared. And and and that's that first party. Who's this Barbie on DH? Finally part. See, we use an equation of motion. Cavil, we have acceleration. We have distance given. So we use the squared equals B not squared. Plus two. Eh? Why? Okay, and and be not be zero. So we simply have that velocity is square root of two times acceleration times, vertical height. And so that would be two times 3.58 times 1.5. This comes out to be 3.28 meters per second square.

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