Answer
$\frac{15}{4}$ $ft-lb$
Work Step by Step
According to Hooke's Law, the force required to maintain a spring stretched $x$ units beyond its natural length (or compressed $x$ units less than its natural length) is proportional to $x$, that is,
$f(x)$ = $kx$
Here, the amount stretched is $4$ $in.$ = $\frac{1}{3}$ $ft$ and natural length to $6$ $in.$ = $\frac{1}{2}$ $ft$ beyond its natural length. Therefore
$10 = \dfrac{1}{3} k\Rightarrow k=30\Rightarrow f(x)=30x$.
The work is
$W$ = $\int_a^b f(x)dx$=$\int_0^{1/2}{30x}dx$ = $[15x^{2}]|_0^{1/2}$ = $\frac{15}{4}$ $ft-lb$