We have 10 springs. Originally, each is $0.5 \mathrm{~m}$ long, with a spring constant of $200 \mathrm{~N} / \mathrm{m}$. We hang a $2 \mathrm{~kg}$ mass on each spring, then hang the springs with the masses in a row. What is the length of the resulting chain? (Neglect the mass of the springs.)
Let's first calculate the elongation of the $i$-th spring from the bottom! For this spring, the weight of $i$ masses of $m=2 \mathrm{~kg}$ acts, that is, $F_{i}=i m g$ force (since the mass of the springs is not to be taken into account according to the problem statement). If we assume that the elongation of the springs is within the elasticity limit, then the elongation $\Delta x_{i}$ of the $i$-th spring can be calculated from the relationship $F_{i}=D \cdot \Delta x_{i}$, where $F_{i}$ is the force acting on the spring, and $D$ is the spring's directional force. From this,

$$
\Delta x_{i}=\frac{F_{i}}{D}=\frac{i m g}{D}
$$

The total elongation of the 10 springs is therefore

$$
\Delta x=\sum_{i=1}^{10} \frac{i m g}{D}=\frac{55 m g}{D}
$$

The total length of the constructed chain is thus

$$
l=\Delta x+10 l_{0}=55 \mathrm{mg} / D+10 l_{0}=10.5 \mathrm{~m}
$$

where $l_{0}$ is the natural length of the springs. (Since the size of the mass $m$ is not given in the problem, we can assume that the length of the mass $m$ is included in the natural length of the spring.)