A uniform solid cylinder of mass 3 kg and radius R lies on a rough horizontal floor with its axis horizontal. A light horizontal spring of stiffness…

Rolling without slipping couples translation and rotation. With friction $f$ and axle acceleration $a$, $Ma=-kx+f$ and the torque about the axis gives $f=-\tfrac12Ma$ (friction opposes the acceleration), hence $\tfrac32Ma=-kx$ and $\omega^2=\dfrac{2k}{3M}=\dfrac{400}{9}$, so $\omega=\tfrac{20}{3}\approx6.7\ \mathrm{rad\,s^{-1}}$. The value $8.2\ \mathrm{rad\,s^{-1}}$ is $\sqrt{k/M}$, which ignores rotation, so (A) is wrong. The friction force is largest where $a$ is largest: $a_{max}=\omega^2A=4.44\ \mathrm{m\,s^{-2}}$ so $f_{max}=\tfrac12Ma_{max}=6.67\ \mathrm{N}$, making (B) correct. The normal force is $Mg=30\ \mathrm{N}$, so rolling needs $\mu_s\ge6.67/30=0.222$; $0.20$ is not enough, so (C) is wrong. Total energy is $\tfrac12kA^2=1\ \mathrm{J}$; at the mean position it is all kinetic with rotational part $\tfrac14Mv^2$ against translational $\tfrac12Mv^2$, so one third, $\tfrac13\ \mathrm{J}$, is rotational and (D) is correct. Correct options: B and D.