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(A) When two springs with stiffness constants $k_1$ and $k_2$ are connected in series, the equivalent spring constant $K_{	ext{eq}}$ is given by the

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$\frac{1}{2}\left(\frac{k_1 k_2}{k_1+k_2}\right) \Delta^2$

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(A) When two springs with stiffness constants $k_1$ and $k_2$ are connected in series, the equivalent spring constant $K_{	ext{eq}}$ is given by the formula: $\frac{1}{K_{	ext{eq}}} = \frac{1}{k_1} + \frac{1}{k_2} = \frac{k_1 + k_2}{k_1 k_2}$ Therefore, $K_{	ext{eq}} = \frac{k_1 k_2}{k_1 + k_2}$. The work done $W$ to stretch a spring system by a total displacement $\Delta$ is equal to the potential energy stored in the equivalent spring: $W = \frac{1}{2} K_{	ext{eq}} \Delta^2$ Substituting the value of $K_{	ext{eq}}$: $W = \frac{1}{2} \left( \frac{k_1 k_2}{k_1 + k_2} ight) \Delta^2$.

$A$ mechanical system consists of two springs with stiffness coefficients $k_1$ and $k_2$ connected in series. The minimum work to be performed on the system to stretch it by a total displacement $\Delta$ is

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