8. (10 points) There are two light springs of equal length but with different stiffnesses. The springs were placed vertically one on top of the other. A load of mass \( m = 3 \) kg was placed on top. As a result, the construction was compressed by \( x_{1} = 40 \) cm. Then the springs were placed side by side and the same load was placed on top again. In this case, the construction was compressed by \( x_{2} = 7.5 \) cm. Find the stiffnesses of the springs. The acceleration due to gravity \( g = 10 \mathrm{m} / \mathrm{s}^{2} \).
Answer: $k_{1}=300 \frac{H}{M}$ and $k_{2}=100 \frac{H}{M}$

Solution. In the first situation, the stiffness of the obtained structure is:

$k_{0}=\frac{k_{1} k_{2}}{k_{1}+k_{2}}$.

The equilibrium condition in this case is: $k_{0} x_{1}=m g$.

As a result: $\frac{k_{1} k_{2}}{k_{1}+k_{2}}=\frac{m g}{x_{1}}=75 \frac{H}{M}$.

In the second situation, the stiffness of the obtained structure is: $k_{\kappa}=k_{1}+k_{2}$ (2 points) The equilibrium condition in this case is: $k_{k} x_{2}=m g$.

As a result: $k_{1}+k_{2}=\frac{m g}{x_{2}}=400 \frac{H}{M}$.

Solving this system of equations, we get: $k_{1}=300 \frac{H}{M}$ and $k_{2}=100 \frac{H}{M}$.