- Thread starter
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#### blazerqb11

##### New member

- Feb 3, 2014

- 3

Example 1.

A cable that weighs 8 lb/ft is used to lift 650 lb of coal up a mine shaft 700 ft deep. Find the work done.

Example 2:

A cylindrically shaped tube has a circular base with a diameter of 2 inches and a height of 12 inches. The bottom of the tube is closed. The tube contains a liquid which is 3 inches deep and has a weight density of 62 lbs per ft

^{3}. What is the work done in pumping the liquid to the top of the tube?

Example 3:

A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 m/s

^{2}for g. Use 1000 kg/m

^{3}as the weight density of water.)

In the first example you find an equation for the force that is done and then integrate with bounds over the distance moved.

In the second example however, you find an equation for the distance a layer is moved, with a constant force multiplied in, and you integrate with bounds from one end of the substance to the other.

The third example seems to combine these two issues into one problem. In this case, the force equation and and the distance equation are multiplied together, but the bounds of integration are still over the substance moved and not the distance.

If someone could explain the differences in these problems, and what exactly the integration is adding up, I would much appreciate it. Specifically I would like to know why the first one integrates with bounds over the distance moved, and doesn't seem to include an equation for the distance in the integral (I'm guessing these are related, but I can't quite put my finger on exactly why), and why the second two examples integrate over the bounds of the substance moved instead.