## Calculus (3rd Edition)

$9$ inches
The work required to compress the spring beyond equilibrium can be calculated as: $\text{Work}, W= \int_{m}^{n} kx \ dx$; where $k$ is the spring constant. and $\dfrac{1}{2} k(1.5)^2=18 \implies k =16\ lb/ft$ Now, $F=k x \implies 12 =16 x$ or, $x=0.75 \ ft$ Thus, a weight of $12 \ lb$ will stretch the spring by $9$ inches.