When a solid cylinder rolls on a horizontal surface, its total kinetic energy is the sum of its translational and rotational kinetic energies. $K = K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$ For a solid cylinder, the moment of inertia about its axis of symmetry is $I = \frac{1}{2}mr^2$. In pure rolling, $v = r\omega$. $K = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\left(\frac{v}{r}\right)^2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$ When the cylinder collides with the spring and comes to rest at maximum compression $x$, its entire kinetic energy is converted into the elastic potential energy of the spring. According to the law of conservation of mechanical energy: $\frac{1}{2}kx^2 = \frac{3}{4}mv^2$ $x^2 = \frac{3mv^2}{2k}$ $x = \sqrt{\frac{3mv^2}{2k}}$ Substitute the given values: $m = 3 \text{ kg}$, $v = 4 \text{ m/s}$, and $k = 200 \text{ N/m}$: $x = \sqrt{\frac{3 \times 3 \times (4)^2}{2 \times 200}} = \sqrt{\frac{9 \times 16}{400}} = \sqrt{\frac{144}{400}} = \frac{12}{20} = 0.6 \text{ m}$ The maximum compression produced in the spring is $0.6 \text{ m}$.