Question

6. A force of 20 lb stretches a spring from a natural length of 7 inches to a length of 12 inches. How much work was performed in stretching the spring to this length?

Answer 1

$$F=k\Delta x$$

$$k=\dfrac{F}{x_2-x_1}$$

$$=\dfrac{20\ lb}{12\ in-7\ in}=4\ lb/in$$

$$W=\dfrac{k(\Delta x)^2}{2}$$

$$=\dfrac{4\ lb/in(12\ in-7\ in)^2}{2}=50\ lb\cdot in$$