A swimming pool has the shape of a rectangular prism with abase that measures 30 ft\operatorname{ft} by 20 ft\operatorname{ft} and is 5 ft\operatorname{ft} deep. The top of the pool is 1 ft\operatorname{ft} above the surface of the water. How much work is required to pump all the water out? Assume the density of water is 62.4 lb\operatorname{lb}/ft3ft^3.

449280 ft ⁡ ∙ lb ⁡ s \operatorname{ft}\bullet\operatorname{lb}s

A swimming pool has the shape of a rectangular prism with abase that measures 30 ft \operatorname{ft} by 20 ft \operatorname{ft} and is 5 ft \operatorname{ft} deep. The top of the pool is 1 ft \operatorname{ft} above the surface of the water. How much work is required to pump all the water out? Assume the density of water is 62.4 lb \operatorname{lb} / f t 3 ft^3 .

this problem, we have a 60 foot cable that is attached to a cylinder that is attached to a winch. We're told the cable weighs 300 pounds, and we wanna know how much work is needed to wind 20 feet of the cable onto the cylinder using the winch. And we're given a hint here. It says divide the cable weight by the cable length to get the density. So here's what's going on here. So we have this kind of building. We'll start we'll just draw a building here to to give some reference to what's happening. And we have this cable that is going off the building. This cable is attached to some kind of cylinder at the top here. It's like looking at a front view of of a cylinder. Right? So we're looking at this front view of a cylinder. There's cable attached to it, and the cable is 60 feet long. Now what's gonna happen is we're eventually going to take this cable and wind it up 20 feet. So we're gonna wind it up to about that point, we'll say. This would mean that we have 20 feet, that has the cable has been winded up, so that's gonna be right there. So we get 20 feet right there. And what we want to know is how much work it's gonna take to do this. Now it's a little bit tricky to just do this right off the bat because we do have an equation for work, which is going to be the integral from zero to h of our weight density, which is gonna be in weight per length multiplied by the length function with all with respect to y. We've seen this equation before. This is how we can find the work for our lifting problem. But it's tricky for this problem because notice we're not given the weight per length anywhere in the problem up here. But there's a way that we can find the weight per length. The weight per length is just going to be the weight that we have right here, the cable weight, which is 300 pounds, and it's going to be divided by the length of the cable, which we're told is 60 feet. So you're just gonna have 300 divided by 60, which is five, and that's gonna be pounds per foot. So this is how we can find our weight per length right here. Now I also need to find the length function l of y, and I can do that by taking my length here of the cable, which is 60 feet, and then subtracting off y, the distance. And this should make sense because as this rope gets lifted farther and farther, less and less of the cable is going to be remaining. So y would be some arbitrary distance that we're lifting it, and we would subtract that from our total length of the cable. So to go ahead and solve this problem, well, we can take everything that we have here and plug it in to this integral. So we have the integral of our weight per length, which we figured out was five, and that's going to be multiplied by our length function, 60 minus y. All of that's integrated with respect to y. Now as for the balance here, we're gonna go from zero to how far we're lifting our cable, which we're told we're lifting at 20 feet. So it's gonna be 20 feet. So this is going to be how we calculate our work. And I can go ahead and take this five, which is a constant, pull it outside the integral right here. So we're gonna go from zero to 20, and then we're gonna have a 60 minus y. Now integrating this, I know that I can just use my rules for integrals. And by the integral of 60, which is going to be 60 y, because with respect to y, so we're going to have 60 y. It's gonna be minus, and then the integral of y is gonna be y squared over two using the power rule, and all of this is gonna be bounded from zero to 20. Now if I plug in zero, every place that we see y is gonna become zero here, which means the whole inside of these brackets will go to zero. So I only need to worry about plugging in my upper bound of 20. So this whole thing is going to be five times 60 times 20 minus 20 squared over two. So you can go ahead and evaluate this first and then multiply everything by five, and the result should come up to 5,000 joules. So 5,000 joules of work is needed to lift this cable 20 feet from its starting position, and that is how you can solve the problem. So this is the solution to the problem. Hope you found this video helpful, and let's move on.

A swimming pool has the shape of a rectangular prism with abase that measures 30 ft ⁡ \operatorname{ft} by 20 ft ⁡ \operatorname{ft} and is 5 ft ⁡ \operatorname{ft} deep. The top of the pool is 1 ft ⁡ \operatorname{ft} above the surface of the water. How much work is required to pump all the water out? Assume the density of water is 62.4 lb ⁡ \operatorname{lb} / f t 3 ft^3 .

449280 ft ⁡ ∙ lb ⁡ s \operatorname{ft}\bullet\operatorname{lb}s

42120 ft ⁡ ∙ lb ⁡ s \operatorname{ft}\bullet\operatorname{lb}s ft ∙ lb s

79560 ft ⁡ ∙ lb ⁡ s \operatorname{ft}\bullet\operatorname{lb}s ft ∙ lb s

280800 ft ⁡ ∙ lb ⁡ s \operatorname{ft}\bullet\operatorname{lb}s ft ∙ lb s

10. Physics Applications of Integrals

Work

Calculus

10. Physics Applications of Integrals

Work

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Multiple Choice

A swimming pool has the shape of a rectangular prism with abase that measures 30 $ft$ by 20 $ft$ and is 5 $ft$ deep. The top of the pool is 1 $ft$ above the surface of the water. How much work is required to pump all the water out? Assume the density of water is 62.4 $lb$ / $f t^{3}$ .

449280 $ft ∙ lb s$

42120 $\operatorname{ft}\bullet\operatorname{lb}s$

79560 $\operatorname{ft}\bullet\operatorname{lb}s$

280800 $\operatorname{ft}\bullet\operatorname{lb}s$

Verified step by step guidance

Step 1: Understand the problem. The work required to pump water out of the pool is calculated using the formula for work: W = ∫ F(y) dy, where F(y) represents the force needed to lift a thin slice of water at height y to the top of the pool. The density of water is given as 62.4 lb/ft³, and the dimensions of the pool are 30 ft by 20 ft by 5 ft.

Step 2: Define the coordinate system. Let y represent the vertical distance from the bottom of the pool. The water extends from y = 0 (bottom of the pool) to y = 5 ft (surface of the water). The top of the pool is at y = 6 ft, so the water must be lifted a distance of (6 - y) ft.

Step 3: Calculate the volume of a thin slice of water. A thin slice of water at height y has a thickness dy and a base area of 30 ft × 20 ft = 600 ft². The volume of this slice is 600 dy ft³.

Step 4: Calculate the weight of the thin slice of water. The weight of the slice is given by multiplying its volume by the density of water: weight = 62.4 × 600 dy = 37440 dy lb.

Step 5: Set up the integral for work. The work required to lift the slice to the top of the pool is the weight of the slice multiplied by the distance it must be lifted: dW = 37440 (6 - y) dy. Integrate this expression from y = 0 to y = 5 to find the total work: W = ∫[0 to 5] 37440 (6 - y) dy.

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