Applying work energy for rigid bodies\[ T_{1}+V_{1}+\int _{0}^{\theta _{2}}Md\theta =T_{2}+V_{2}\] But \(V_{1}=V_{2}\), and let \(F=\mu _{k}P\), where \(P\) is the force pushing down and \(F\) is the friction force, then\begin{align*} \frac{1}{2}I_{cg}\omega _{1}^{2}+\int _{0}^{\theta _{2}}Md\theta & =0\\ \frac{1}{2}\left ( mr^{2}\right ) \omega _{1}^{2}+\int _{0}^{\theta _{2}}-Frd\theta & =0\\ \frac{1}{2}\left ( mr^{2}\right ) \omega _{1}^{2}+P\mu _{k}r\theta _{2} & =0 \end{align*}
But \(\theta _{2}=6\pi \) since \(3\) revolutions, then\[ \frac{1}{2}\left ( mr^{2}\right ) \omega _{1}^{2}-P\mu _{k}r\left ( 6\pi \right ) =0 \] Solving for \(P\)\begin{align*} P & =\frac{\frac{1}{2}\left ( mr^{2}\right ) \omega _{1}^{2}}{\mu _{k}r\left ( 6\pi \right ) }\\ & =\frac{\frac{1}{2}\left ( \left ( 1.1\right ) \left ( 0.3\right ) ^{2}\right ) \left ( 13.3\right ) ^{2}}{\left ( 0.3\right ) \left ( 0.3\right ) \left ( 6\pi \right ) }\\ & =5.161\text{ N} \end{align*}
When the disk is pulled, it gains potential energy \(V_{1}=\frac{1}{2}kx^{2}\) where \(x\) is amount of spring extension from equilibrium, which is \(1.7\) meter in this example. When released, all this energy will be converted to kinetic energy when the disk reaches its original equilibrium position. The final kinetic energy is \(T_{2}=\frac{1}{2}mv_{g}^{2}+\frac{1}{2}I_{cg}\omega ^{2}\). But since the disk rolls without slip, then \(v_{cg}=r\omega \) and \[ T_{2}=\frac{1}{2}mv_{g}^{2}+\frac{1}{2}I_{cg}\frac{v_{cg}^{2}}{r^{2}}\] But \(I_{cg}=\frac{1}{2}mr^{2}\) and the above becomes\begin{align*} T_{2} & =\frac{1}{2}mv_{g}^{2}+\frac{1}{4}mr^{2}\frac{v_{cg}^{2}}{r^{2}}\\ & =\frac{1}{2}mv_{g}^{2}+\frac{1}{4}mv_{cg}^{2}\\ & =\frac{3}{4}mv_{cg}^{2} \end{align*}
Hence equating the initial potential energy to the final kinetic energy we obtain\[ \frac{1}{2}kx^{2}=\frac{3}{4}mv_{g}^{2}\] Solving for \(v_{g}\) gives\begin{align*} v_{cg}^{2} & =\frac{\frac{1}{2}kx^{2}}{\frac{3}{4}m}\\ & =\frac{2}{3}\frac{kx^{2}}{m}\\ & =\frac{2}{3}\frac{\left ( 20\right ) \left ( 1.7\right ) ^{2}}{14}\\ & =2.7524 \end{align*}
Therefore\begin{align*} v_{cg} & =\sqrt{2.7524}\\ & =1.659\text{ m/s} \end{align*}