Single Variable Calculus, Chapter 6, 6.4, Section 6.4, Problem 12

Suppose that $6J$ of work is required to stretch a spring from $10cm$ to $12cm$ and another $10J$ is needed to stretch it from $12cm$ to $14cm$, what is the natural length of the spring?

Let $L$ be the natural length of the spring in meters.

So,


$\begin{equation} \begin{aligned} & W = \int^b_a f(x) dx \\ \\ & W = \int^b_a kx dx \\ \\ & 6 = \int^{0.12 - L}_{0.10 - L} kx dx; \text{ recall that $1m = 100 cm$} \\ \\ & 6 = k \left[ \frac{x^2}{2} \right]^{0.12 - L}_{0.10 - L} \\ \\ & 6 = \frac{k}{2} \left[(0.12 - L)^2 - (0.10 - L)^2\right] \\ \\ & 12 = k [0.0144 - 0.24 L + \cancel{L^2} - 0.01 + 0.20 L - \cancel{L^2}] \\ \\ & k [0.0044 - 0.04 L] \qquad \text{ Equation 1} \end{aligned} \end{equation}$



Similarly, from the other point,


$\begin{equation} \begin{aligned} & W = \int^b_a kx dx \\ \\ & 10 \int^{0.14 - L}_{0.12 - L} kx dx \\ \\ & 10 = k \left[ \frac{x^2}{2} \right]^{0.14 - L}_{0.12 - L} \\ \\ & 10 = \frac{k}{2} [(0.14 L)^2 - (0.12 - L)^2] \\ \\ & 20 = k \left[0.0196 - 0.28L + L^2 - 0.0144 + 0.24L - L^2\right] \\ \\ & 20 = k[0.0052 - 0.04L]\qquad \text{ Equation 2} \end{aligned} \end{equation}$


Use equations 1 and 2 to solve for $k$ and $L$ simultaneously,


$\begin{equation} \begin{aligned} k =& 10,000 \\ \\ L =& \frac{32}{400} m \text{ or } 8cm \end{aligned} \end{equation}$



Therefore, the natural length of the spring is $L = 8cm$.