Single Variable Calculus, Chapter 6, 6.4, Section 6.4, Problem 8

Suppose that a spring has a natural length of $20 cm$. If a $25-N$ force is required to keep it stretched to a length of $30 cm$, how much work is required to stretch it from $20 cm$ to $25 cm$.

Return from Hooke's Law:

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

So,


$
\begin{equation}
\begin{aligned}

& 25N = k (30 cm - 20 cm) \left( \frac{1m}{100 cm} \right)
\\
\\
& 25N = k(0.1)m
\\
\\
& k = 250 \frac{N}{m}

\end{aligned}
\end{equation}
$


Thus,


$
\begin{equation}
\begin{aligned}

& f(x) = 250x
\\
\\
& \text{then,}
\\
\\
& W = \int^{0.05}_0 250x dx
\\
\\
& W = \left[ \frac{250x^2}{2} \right]^{0.05}_0
\\
\\
& W = \frac{5}{16} \text{ Joules}

\end{aligned}
\end{equation}
$