Question

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?

Answer 1

$F_1=k\Delta x_1=k\cdot \frac{L}{10}\implies k=\frac{10F_1}{L}=\frac{50}{L},$

$F_2=k\Delta x_2=kL=\frac{50}{L}\cdot L=50~\text{lb}.$