A spring has a natural length of 2 ft, and a force of a 15 lb is required to hold it compressed at a length of 18 in. How much work is done in stretching this spring from its natural length to a length of 3 ft?

Working:

$\displaystyle 18 in=\frac{18}{12} ft=1.5 ft$

x=compression=2-1.5=0.5 ft

F=kx

15=k*0.5

k=30 lb/ft

$\displaystyle W=\frac{1}{2}kx_2^2-\frac{1}{2}kx_1^2$

= $\displaystyle \frac{1}{2}k(x_2^2-x_1^2)$

= $\displaystyle \frac{30}{2}(1^2-0^2)$

= 15*1

= 15 lb-ft

Edit:Never mind; I solved it myself.