Single Variable Calculus, Chapter 6, 6.4, Section 6.4, Problem 10

If $12ft-lb$ is the required work to stretch a spring $1$ ft beyond its natural length, how much work is needed to sketch $9$ inches beyond its vertical length?

Recall from Hooke's Law,

$f(x) = kx$ where $x$ is the maximum elongated length of the spring and $k$ is the spring constant


$\begin{equation} \begin{aligned} & W = \int^b_a f(x) dx \\ \\ & W = \int^b_a kx dx \\ \\ & W = \int^1_0 kxdx \quad \text{; recall that 9 inches } \left(\frac{1\text{ft}}{12 \text{inches}}\right)\\ \\ \\ & 12 = k \left[ \frac{x^2}{2} \right]^1_0 \\ \\ & 12 = k \left( \left[ \frac{(1)^2}{2} \right] - \left[ \frac{(0)^2}{2} \right] \right) \\ \\ & 12 = k \left( \frac{1}{2} \right) \\ \\ & k = 24 \frac{lb}{ft} \end{aligned} \end{equation}$


Therefore, the required work to stretch the string to $9$ inches is...


$\begin{equation} \begin{aligned} W =& \int^{\frac{3}{4}}_0 kx dx; \text{ recall that } 9 \text{ inches }\left( \frac{1 \text{ft}}{12 \text{inches}} \right) = \frac{3}{4} ft \\ \\ W =& \int^{\frac{3}{4}}_0 24x dx \\ \\ W =& \frac{27}{4} ft -lb \end{aligned} \end{equation}$