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Chapter 6 | Applications of Integration
The Work Required to Stretch or Compress a Spring
Suppose it takes a force of 10 N (in the negative direction) to compress a spring 0.2 m from the equilibrium position. How much work is done to stretch the spring 0.5 m from the equilibrium position?
Solution First find the spring constant, k . When x =−0.2, we know F ( x ) =−10, so F ( x ) = kx −10 = k (−0.2) k = 50 and F ( x ) =50 x . Then, to calculate work, we integrate the force function, obtaining W = ∫ a b F ( x ) dx = ∫ 0 0.5 50 xdx =25 x 2 | 0 0.5 =6.25. The work done to stretch the spring is 6.25 J.
6.25 Suppose it takes a force of 8 lb to stretch a spring 6 in. from the equilibrium position. How much work is done to stretch the spring 1 ft from the equilibrium position?
Work Done in Pumping Consider the work done to pump water (or some other liquid) out of a tank. Pumping problems are a little more complicated than spring problems because many of the calculations depend on the shape and size of the tank. In addition, instead of being concerned about the work done to move a single mass, we are looking at the work done to move a volume of water, and it takes more work to move the water from the bottom of the tank than it does to move the water from the top of the tank. We examine the process in the context of a cylindrical tank, then look at a couple of examples using tanks of different shapes. Assume a cylindrical tank of radius 4 m and height 10 m is filled to a depth of 8 m. How much work does it take to pump all the water over the top edge of the tank? The first thing we need to do is define a frame of reference. We let x represent the vertical distance below the top of the tank. That is, we orient the x -axis vertically, with the origin at the top of the tank and the downward direction being positive (see the following figure).
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