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How much work is done in stretching the spring of Exercise 3 from a length of 10.0 in. to 12.0 in.?

Calculus 2 / BC

Chapter 26

Applications of Integration

Section 6

Other Applications

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This problem we have a continuation of the previous problem where we are asked how much work is done in stretching the spring of exercise three from a length of 10 inches to 12 inches. Where we found that for the spring in exercise three, our spring constant was equal to four. So in this case we'd have that the work done will be the integral from 10.8 or 10 up to 12 of four X. Dx. Which will then be four x squared over two, evaluated from 10 to 12 or two X squared, evaluated from 10 to 12. It will then be two times 12 squared, So two times 1 44 minus 10 squared, so minus 100 Which will then clearly B2 times 44 or 88 pound inches.

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