$\begin{array}{1 1}7.6389ft-lbs\\6.6389ft-lbs\\5.6389ft-lbs\\4.6389ft-lbs\end{array} $

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Answer : 7.6389ft-lbs

We are told that a force of 20lbs is needed to stretch the spring 24in-18in =6in=0.5ft from its natural length.

Then using Hooke's law we have, $20=k(0.5)$

$\Rightarrow k= 40$

Again,using Hooke's law we can see that the force function is $F(x) =40x$

For the limits of the integral we can see that we start with the spring at a length of 21-18=3in or $\large\frac{1}{4}$ feet and we end with a length of 26 - 18=8in or $\large\frac{2}{3}$ feet.

These are then the limits of the integral. The work is then, $W=\int\limits_{\large\frac{1}{4}}^{\large\frac{2}{3}} 40\;x\;dx$

$\;\;\;\;=20 x^2\bigg|_{\Large\frac{1}{4}}^{\Large\frac{2}{3}}$

$\;\;\;\;=\large\frac{275}{36}$

$\;\;\;\;=7.6389\;ft-lbs$

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