6. (15 points) Two springs with stiffnesses of $6 \mathrm{kH} / \mathrm{m}$ and $12 \mathrm{kH} / \mathrm{m}$ are connected in series. How much work is required to stretch this system by 10 cm.
Answer: 20 J.

Solution. Total stiffness: $k=\frac{k_{1} k_{2}}{k_{1}+k_{2}}=\frac{6 \cdot 12}{6+12}=4 \mathrm{kH} / \mathrm{m}$. Work:

$$
A=\Delta W=\frac{k x^{2}}{2}=\frac{4000 \cdot \cdot 0.1^{2}}{2}=20 \text { J. }
$$