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Section 6-6 : Work

2. A spring has a natural length of 18 inches and a force of 20 lbs is required to stretch and hold the spring to a length of 24 inches. What is the work required to stretch the spring from a length of 21 inches to a length of 26 inches?

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Hint : What is the spring constant, \(k\) and the force function?
Start Solution

Let’s start off by finding the spring constant. We are told that a force of 20 lbs is needed to stretch the spring 24 in – 18 in = 6 in = 0.5 ft from its natural length. Then using Hooke’s Law we have,

\[20 = k\left( {0.5} \right)\,\hspace{0.5in} \Rightarrow \hspace{0.5in}k = 40\]

Don’t forget that we want the displacement in feet. Also, don’t forget that the displacement needs to be the displacement from the natural length of the spring.

Again, using Hooke’s Law we can see that the force function is,

\[F\left( x \right) = 40x\] Show Step 2

For the limits of the integral we can see that we start with the spring at a length of 21 in – 18 in = 3 in or \(\frac{1}{4}\) feet and we end with a length of 26 in – 18 in = 8 in or \(\frac{2}{3}\) feet. These are then the limits of the integral (recall that we need the relative distance from the natural length for the limits).

The work is then,

\[W = \int_{{\frac{1}{4}}}^{{\frac{2}{3}}}{{40x\,dx}} = \left. {20{x^2}} \right|_{\frac{1}{4}}^{\frac{2}{3}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{275}}{{36}} = 7.6389\,{\mbox{ft - lbs}}}}\]