Work in foot pounds

Question

A force of 2 pounds is required to hold a spring stretched 0.3 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 0.8 feet beyond its natural length?

Answer 1

Force, F = kx

If F is 2 and x is 0.3, then you can solve for k. Then you can use that value of k to find the Work for x=0.8, using

Work, W = ∫₀^0.8kxdx = kx²/2 |₀^0.8