Two springs with stiffnesses k_1 and k_2 are connected in…

Two springs with stiffnesses $k_1$ and $k_2$ are connected in series. What is the minimum work needed to stretch the pair by $\Delta x$?

Explanation of Method

Work to stretch two springs in series.

Step 1: Equivalent stiffness

Both springs carry the same force, and their extensions add: $\Delta x = \frac{F}{k_1} + \frac{F}{k_2} \;\Longrightarrow\; k = \frac{k_1k_2}{k_1 + k_2}$.

Step 2: Work

\[W = \frac12\,\frac{k_1k_2}{k_1 + k_2}\,\Delta x^2\]

$k_1$ $k_2$ $F$ $x_1 = F/k_1$ $x_2 = F/k_2$ $\Delta x = x_1 + x_2 = F\left(\tfrac{1}{k_1} + \tfrac{1}{k_2}\right)$

In series both springs carry the same force, and their extensions add. The pair behaves like one softer spring.

Result

\[\boxed{\,W = \frac12\,\frac{k_1k_2}{k_1 + k_2}\,\Delta x^2\,}\]

Remark

Check. The original result is correct. "Minimum" means stretching slowly, without giving the spring ends kinetic energy.