Question

if the work required to stretch a spring 1 ft beyond its natural length is 12 ft-lb, how much work is needed to stretch it 9 in beyond its natural length?

Answer 1

\[W=F\cdot d~~\iff~~12~\text{lb}=F\]
It takes 12 N of force to stretch the spring 1 ft. Hooke's Law says
\[F=kx~~\iff~~12~\text{lb}=k\cdot(1~\text{ft})~~\iff~k=12~\frac{\text{lb}}{\text{ft}}\]
Find the force it takes the stretch the spring 9 inches:
\[F=\left(12~\frac{\text{lb}}{\text{ft}}\right)\cdot\left(\frac{9}{12}~\text{ft}\right)=9~\text{lb}\]
The work would then be
\[W=(9~\text{lb})\cdot\left(\frac{9}{12}~\text{ft}\right)=\frac{81}{12}~\text{ft-lb}\]