6. (15 points) Two springs with stiffnesses of 3 kN/m and $6 \mathrm{kN} / \mathrm{m}$ are connected in series. How much work is required to stretch this system by 5 cm.
Answer: 2.5 J.

Solution. The total stiffness $k=\frac{k_{1} k_{2}}{k_{1}+k_{2}}=\frac{3 \cdot 6}{3+6}=2 \mathrm{kH} /$ m. Work:

$$
A=\Delta W=\frac{k x^{2}}{2}=\frac{2000 \cdot 0.05^{2}}{2}=2.5 \text { J. }
$$

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