A uniform solid cylinder of mass 2 kg rolls without slipping on a rough horizontal floor with centre speed 4 m s^-1. A light spring of stiffness 300…

The floor friction does no work because the contact point is instantaneously at rest, and the spring is conservative, so mechanical energy is conserved, but the energy of the cylinder is translational plus rotational. For a solid cylinder $I=\tfrac12 mR^2$ and rolling gives $\omega=v/R$, so $K=\tfrac12mv^2+\tfrac12 I\omega^2=\tfrac34 mv^2=\tfrac34(2)(16)=24\ \mathrm{J}$. At maximum compression the cylinder is momentarily at rest (and so is not rotating), so all of this goes into the spring: $\tfrac12kx^2=24$, giving $x^2=48/300=0.16$ and $x=0.40\ \mathrm{m}$. Using only the translational energy $\tfrac12mv^2=16\ \mathrm{J}$ gives $0.33\ \mathrm{m}$; the other options do not correspond to the correct energy balance.