Energy Method for Spring-Pulley Constraints
Let the tension in the single continuous cord be $T$. We examine the forces acting on the block and the springs.
On the Block: The diagram indicates that the block (and its attached pulley system) is pulled back by 3 segments of the rope. Thus, the total resisting force from the rope is $3T$. Since the block is pulled by force $F$ in equilibrium (or considering work done):
$$F = 3T \implies T = \frac{F}{3}$$
On the Springs: The cord loops around the pulley attached to each spring. This means each spring pulley is pulled by two segments of the rope. Therefore, the force stretching each spring is:
$$F_{spring} = 2T$$
We apply the Work-Energy Theorem. The work done by the external force $F$ in moving the block by a distance $X$ is stored as potential energy in the two springs.
$$W = U_{total}$$
Work done by Force $F$ (assuming gradual loading linear average or direct energy balance):
$$W = \frac{1}{2} F X$$
Potential Energy in Springs:
There are 2 identical springs with stiffness $k$. The extension in each spring is $x_s$.
$$kx_s = F_{spring} = 2T$$
$$x_s = \frac{2T}{k}$$
Total Potential Energy:
$$U = 2 \times \left( \frac{1}{2} k x_s^2 \right) = k \left( \frac{2T}{k} \right)^2 = \frac{4T^2}{k}$$
Equating Work and Energy:
$$\frac{1}{2} F X = \frac{4T^2}{k}$$
Substitute $T = F/3$:
$$\frac{1}{2} F X = \frac{4}{k} \left( \frac{F}{3} \right)^2$$
$$\frac{1}{2} F X = \frac{4}{k} \cdot \frac{F^2}{9}$$
$$\frac{1}{2} X = \frac{4F}{9k}$$
$$X = \frac{8F}{9k}$$