We analyze the equilibrium position of the block of mass $m = 8.0 \text{ kg}$. The system consists of two springs with stiffness $k_1 = 100 \text{ N/m}$ and $k_2 = 200 \text{ N/m}$. Let $y$ be the vertical descent of the block, $x_1$ be the extension in spring $k_1$, and $x_2$ be the extension in spring $k_2$.

Using the Principle of Virtual Work, the total work done by all forces for a virtual displacement must be zero. The work done by gravity is balanced by the potential energy stored in the springs. The equation for virtual work is:

This implies that at equilibrium, the decrease in gravitational potential energy equals the increase in elastic potential energy: $$ mg \delta y = k_1 x_1 \delta x_1 + k_2 x_2 \delta x_2 $$

Let $T$ be the tension in the string supporting the lower pulley (which holds the mass). From the free-body diagram of the mass $m$:

Analyzing the connections to the springs based on the string geometry shown in the problem, we identify the force relations:

• For the right spring ($k_2$), the string tension acts directly: $$ k_2 x_2 = T = \frac{mg}{2} $$
• For the left spring ($k_1$), the geometry of the pulley system (as indicated by the effective tension distribution) results in a force relation: $$ k_1 x_1 = \frac{T}{2} = \frac{mg}{4} $$

The kinematic constraint relating the displacements is derived from these force relations (using $Work = Force \times Displacement$):

Now, we solve for $x_1$ and $x_2$ in terms of $mg$ and substitute into the displacement equation.

Step A: Find $x_2$

Step B: Find $x_1$

Step C: Calculate Total Descent $y$

Substituting $x_1$ and $x_2$ into the constraint equation:

Given values:

• $m = 8.0 \text{ kg}$
• $g = 10 \text{ m/s}^2$
• $k_1 = 100 \text{ N/m}$
• $k_2 = 200 \text{ N/m}$