Push it away

The distance between the blocks is $x = 2R\cos\theta$ , and hence $\dot{x} \; = \; -2R\sin\theta \dot{\theta} \hspace{2cm} \ddot{x} \; = \; -2R\sin\theta \ddot{\theta} - 2R\cos\theta \dot{\theta}^2$ The kinetic energy of the plank is $\tfrac12m\dot{x}^2$ . Since the moment of inertia of the cylinder about the point of contact with the rough block is $\tfrac12mR^2 + mR^2 = \tfrac32mR^2$ , the kinetic energy of the cylinder is $\tfrac34mR^2\dot{\theta}^2$ . Thus conservation of energy tells us that $\tfrac12m\dot{x}^2 + \tfrac34mR^2\dot{\theta}^2 + mgR\sin\theta \; = \; mgR\sin\theta_0$ and hence $(3 + 8\sin^2\theta)\dot{\theta}^2 \; =\; \tfrac{4g}{R}(\sin\theta_0 - \sin\theta)$ where $\theta_0 = 37^\circ$ is the original angle. Thus $2(3 + 8\sin^2\theta)\ddot{\theta} + 16\sin\theta \cos\theta \dot{\theta}^2 \; = \; -\tfrac{4g}{R}\cos\theta$ Initially we have $\theta = \theta_0$ , $\dot{\theta} = 0$ , and hence $\ddot{\theta} \; = \; -\frac{2g\cos\theta_0}{R(3 + 8\sin^2\theta_0)} \hspace{2cm} \ddot{x} \; = \; \frac{4g\sin\theta_0\cos\theta_0}{3 + 8 \sin^2\theta_0}$ initially. Since $m\ddot{x} = N\cos\theta_0$ , where $N$ is the normal reaction between the smooth plank and the cylinder, we see that the initial value of $N$ is $\frac{4mg\sin\theta_0}{3 + 8\sin^2\theta_0} \; = \; 20.00$ making the answer $\boxed{20\; \mathrm{N}}$ .

5kg * 10 m/s = 50N
37 is roughly 40 = 0.4
50N * 0.4 = 20N

Sry for no format at all it didn't work :(