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Calculate the work required to lift a $10-\mathrm{m}$ chain over the side of a building (Figure 13$)$ Assume that the chain has a density of 8 $\mathrm{kg} / \mathrm{m}$ . Hint: Break up the chain into $N$ segments, estimate the work performed on a segment, and compute the limit as $N \rightarrow \infty$ as an integral.

3920$J$

Calculus 1 / AB

Calculus 2 / BC

Chapter 6

APPLICATIONS OF THE INTEGRAL

Section 5

Work and Energy Preparing for the AP Exam

Integrals

Integration

Applications of Integration

Campbell University

Harvey Mudd College

Baylor University

Idaho State University

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whereas to calculate. I'm not a work required to lift a 10 meter train over onto the side of the building. So what we're gonna do is we're going to divide this chain into a multiple, multiple, tiny segment, and each segment is going to have a length of built A why? Okay, so the length of each segment is going to be equal to Delta White. Next. We're given that the density of the chain is eight kilometers per meter. So this means the mass is going to be the density times the length, which is going to be eight. Trying still tell wife, which means next we can find our force. The force is going to be the mass times gravity because we're lifting it up, which means we're working against gravity. So force is going to be eight times 9.8 times built a wife case. Well, the last thing we want to find is how much work is required to lift that one tiny segment up onto the roof. So if our segment is, let's say here, then the amount of distance that it has to go is going to be. Let's call that why so that means the work is going to be equal to force times. Why. So now? If we do that for each tiny segment of the chain and then we sum it all up, then that's going to converge to an integral So that tells us that the total work is going to be equal to the integral of lifetimes. Why is that? I'll be eight times 9.8, which is 78 point four times why in Delta Y is now D y. So the limits of integration is, um, didn't like the distance we have toe hold. The train from the smallest isn't the longest distance. So the first segment it would be right at the rooftop so that that's going to be a distance of zero. And then the furthest distance would be the last segment down here. And that would have to go a distance of 10 since the changes 10 meters long, so the upper limit is going to be 10. Now, this is going to be 78.4 times half. Why squared validated from 0 to 10. So this is going to be equal to four times 50. So your answer is going to be 3920. Jules

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