(a) Let us indicate the forces and their points of application for the cylinder. Choosing the positive direction for $x$ and $\varphi$ as shown in the figure, we write the equation of motion of the cylinder axis and the equation of moments in the C.M. frame relative to that axis
i.e. from equation $F_{x}=m w_{c}$ and $N_{z}=I_{c} \beta_{z}$
As there is no slipping of thread on the cylinder $w_{c}=\beta R$
From these three equations $T=\frac{m g}{6}=13 \mathrm{~N}, \beta=\frac{2}{5} \frac{g}{R}=5 \times 10^{2} \mathrm{rad} / \mathrm{s}^{2}$
(b) we have $\beta=\frac{2}{3} \frac{g}{R}$
So, $w_{c}=\frac{2}{3} g>0$ or, in vector form $\vec{w}_{c}=\frac{2}{3} \vec{g}$
$P=\vec{F} \cdot \vec{v}=\vec{F} \cdot\left(\vec{w}_{c} t\right)$
$=m \vec{g} \cdot\left(\frac{2}{3} \overrightarrow{g t}\right)=\frac{2}{3} m g^{2} t$