Rod Pushing Block

The moment of inertia of the rod about the hinge is $\tfrac13 \times 1 \times 1^2 = \tfrac13$ , and so the KE of the rod is $\tfrac16\dot\theta^2$ , where $\theta$ is the angle between the rod and the horizontal. If $x$ is the distance from the left-hand edge of the block to the hinge, then the KE of the block is $\tfrac12 \times 1 \times \dot{x}^2 = \tfrac12\dot{x}^2$ . Conservation of energy tells us that $\tfrac16\dot{\theta}^2 + \tfrac12\dot{x}^2 + \tfrac12g\sin\theta \; = \; \tfrac{g}{2\sqrt{2}}$ At any time, the proportion of KE that is associated with the block is $R \; = \; \frac{\tfrac12\dot{x}^2}{\tfrac16\dot{\theta}^2 + \tfrac12\dot{x}^2} \; =\; \frac{3\dot{x}^2}{\dot{\theta}^2 + 3\dot{x}^2}$ But $\tan\theta = \tfrac{1}{3x}$ , and so $\sec^2\theta \dot{\theta} \; = \; -\tfrac{1}{3x^2}\dot{x}$ , so that $R \; =\; \frac{27x^4}{27x^4 + \cos^4\theta}$ At the moment we are interested, $x = \cos\theta = \tfrac{2\sqrt{2}}{3}$ , and so $R = \tfrac{27}{28}$ , this makes the answer $\boxed{96.428571423...}$ .

Conservation of energy is not needed to solve this problem. However, conservation of energy, together with the relationship between $\dot{x}$ and $\dot{\theta}$ , could be used to determine a differential equation for $x$ , and hence establish the full dynamics of the problem.