How much work (in foot-pounds) is done in stretching the spring from its natural length to 1.1 feet beyond its natural length? foot-pounds

Question

A force of 3 pounds is required to hold a spring stretched 0.2 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 1.1 feet beyond its natural length? foot-pounds

Answer 1

F=k(x₁ - x₀); k = F/(x₁ - x₀) = 3/0.2 = 15lb/foot;

W = k(x₂ - x₀)²/2 = 15·1.1²/2lb·feet = 9.075 lb·feet