Emptying a water trough A water trough has a semicircular cros...

Question

Emptying a water trough A water trough has a semicircular cross section with a radius of 0.25 m and a length of 3 m (see figure).

a. How much work is required to pump the water out of the trough (to the level of the top of the trough) when it is full?

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Accepted Answer

Hello. In this video, we are told that a trench filled with water has a triangular cross section with a base of 1 m, a height of 0.5 m, and a length of 4 m. The trench is emptied by pumping the water up to the top of the trench and out of it. How much work is required to empty the trench, and we want to use that row is equal to 1000 kg per meter's cube, and G is the gravitational constant. Now, what we need to do is we need to go ahead and solve for an equation for work. Now, in this case here, we have that the differential of work is defined as row multiplied by G, multiplied by the height of the tank, multiplied by the area with respect toy of the tank, D Y. Now, in this case here, row and G are given to us as constant values, so we don't need to worry about this. But what we need to do is we need to go ahead and define the height of the tank, and define the area of the tank. Now, the area of the tank is going to be defined as the length multiplied by the width of the tank in terms of Y. Now, if we think, take a look at the width of the tank, the width of the tank in terms of Y is proportional to the height Y, and that is equal to the base multiplied by the height. Now in this case here, In this case here, if we simplify, we know that the base of the side of the tank is defined as 1 m, and the height of the tank is 0.5. So we get that width in terms of Y, divided by Y is equal to 1 divided by 0.5. This is going to simplify to be W of Y divided by Y is equal to 2, and by multiplying both sides of the equation by Y, we get that the width of the tank in terms of Y is equal to 2 Y. And so by taking this value and plugging it back into the area of the tank, we get that the area is equal to the length times the width, and again the length of the tank is given to us as 4 m. So, area is equal to 4, multiplied by 2Y, which is going to be 8Y. Now, in this case here, we know we need to come up with an equation for height in terms of Y. Now, we know that the height of the tank is going to be 0.5, but it's with respect to some height Y. So in this case here, height is going to be defined as 0.5 minus Y. And so now that we have the height of the tank and the area of the tank, let's go ahead and plug this into the work differential. So by plugging this in, we get that the differential of work is equal to row, multiplied by G, multiplied by the quantity, 0.5 minus Y, multiplied by 8 Y D Y. And so now what we're going to do is we're going to simplify this differential. We will distribute 8 Y into 0.5 minus Y, and that gives us DW is equal to row, multiplied by G, multiplied by the quantity for Y. Minus 8 Y 2 D Y. Now, in order to solve for an equation of work, what we need to do is we need to remove the differentials. In order to remove the differentials, we will integrate both sides of this equation with respect to the variables. And by doing this, we get that the left hand side is going to simplify to be work, and that is equal to the constants row multiplied by G. Multiplied by a definite integral, and in this case here, because we're integrating with respect to Y, we are going to integrate from 0 to 0.5. Of the quantity for Y minus 8 Y squared, D Y. Now we can solve for work by just integrating term by turn. So we get row multiplied by G, multiplied by the quantity, 2 Y squared, minus 8/3 Y cubed, and this will be evaluated from 0 to 0.5. Now, if we were to solve or evaluate from the bounce, the antiderivative on the bounce from 0 to 0.5, we get that we get row multiplied by G multiplied by 16. And now if we plug in the values of row and G, we are left with 1/6 multiplied by 1000 multiplied by 9.8, and this is going to simplify to be 4900 divided by 3 joules. This is going to be the force that we are looking for, and the overall solution to this problem is going to be B. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

Emptying a water trough A water trough has a semicircular cross section with a radius of 0.25 m and a length of 3 m (see figure).
a. How much work is required to pump the water out of the trough (to the level of the top of the trough) when it is full?

Key Concepts

Work Done by a Variable Force

Volume and Cross-Sectional Area of a Semicircle

Density and Weight of Water

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