If a force of 5 lb produces a stretch of one-tenth the natural length L ft of a spring. How much work will be done in stretching the spring to double its natural length?
F1=kΔx1=k⋅L10 ⟹ k=10F1L=50L,F_1=k\Delta x_1=k\cdot \frac{L}{10}\implies k=\frac{10F_1}{L}=\frac{50}{L},F1=kΔx1=k⋅10L⟹k=L10F1=L50,
F2=kΔx2=kL=50L⋅L=50 lb.F_2=k\Delta x_2=kL=\frac{50}{L}\cdot L=50~\text{lb}.F2=kΔx2=kL=L50⋅L=50 lb.
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