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The spring of a spring balance is 8.0 in. long when there is no weight on the balance, and it is 9.5 in. long with 6.0 lb hung from the balance. How much work is done in stretching it from 8.0 in. to a length of 10.0 in?

Calculus 2 / BC

Chapter 26

Applications of Integration

Section 6

Other Applications

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For this problem we are told that the spring of a spring balance is 8" long when there is no weight on the balance and it is 9.5" long when £6 is hung from the balance. Were then asked how much work is done in stretching it from 8" to a length of 10 in. So to begin we have hooks law which gives us that F of X would be equal to K. X. Where X is the displacement from equilibrium. So in this case we would have that. Let's see here. So you'd have that 6.0 or £6.8 is the amount of force needed to stretch this? A total distance of 1.5 inches. So six point oh equals K. Times 1.5. Which then means that K must be equal to 6.0 divided by 1.5 which would then be four. So we have that F of X equals four X. Then the work done in this case and we'll be the integral of f of X from a total stretch of zero. It's starting on stretched up to let's see here, we're going from eight inches to a length of 10 inches. So we're stretching it by a total of two inches two inches rather. So would be the integral from 0 to 2 of four. X. Dx which will be four X squared over two, evaluated from 0 to 2 or simply two X squared evaluated from 0 to 2 which would then be two times two squared or two times four giving us eight. So the total amount of work done is um £8 inches.

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