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 $L = 8cm$.