Force, stretching, and natural length???

Question

A force of 3 pounds is required to hold a spring stretched 0.5 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 0.7 feet beyond its natural length? 

Please write answer as well as any work to get to the answer I'm very lost! T-T

Answer 1

The spring Force equation is F = -kΔx. So we can plug what we know in and solve for k: 3 = -k(0.5). So k = -6. Now to find work, just integrate this Force on the interval that is being stretched.

∫₀^0.7 6x dx = 3x² |₀^0.7 = 3(0.49) = 1.47

So the work to stretch this spring 0.7 feet is 1.47 foot-pounds.