Answer:
(a) [tex]KE_\text{translational} \approx 3.4 \times 10^2 \; \rm J[/tex].
(b) [tex]KE_\text{rotational} \approx 1.7 \times 10^2 \; \rm J[/tex].
(c) [tex]KE_\text{translational} + KE_\text{rotational} \approx 5.1 \times 10^2\; \rm J[/tex].
Explanation:
The translational kinetic energy of an object with mass [tex]m[/tex] and velocity [tex]v[/tex] is
[tex]\displaystyle \text{KE}_\text{translational} = \frac{1}{2}\, m \cdot v^2[/tex].
In this question, [tex]m = 14.0\; \rm kg[/tex] and [tex]v = 7.0\; \rm m \cdot s^{-1}[/tex].
[tex]\begin{aligned}&\text{KE}_\text{translational}\\ &= \frac{1}{2}\, m \cdot v^2 \\&= \frac{1}{2}\times 14.0 \times 7.0^2 \approx 3.4\times 10^{2}\; \rm J\end{aligned}[/tex].
Let [tex]r[/tex] be the radius of this cylinder. Since this cylinder is rolling without slipping, its angular velocity would be [tex]\displaystyle \omega = \frac{v}{r}[/tex].
The moment of inertia of a cylinder along its central axis is [tex]\displaystyle I = \frac{1}{2}\, m \cdot r^2[/tex].
The rotational kinetic energy of that cylinder would be
[tex]\begin{aligned} & KE_\text{rotational} \\ &= \frac{1}{2}\, I \cdot \omega^2 \\ &= \frac{1}{2} \, \left(\frac{1}{2}\, m \cdot r^2\right) \cdot \left(\frac{v}{r}\right)^2 \end{aligned}[/tex].
Note that the radius of the cylinder, [tex]r[/tex], can be eliminated from this expression.
[tex]\begin{aligned} & KE_\text{rotational} \\\ &= \frac{1}{2} \times \left(\frac{1}{2} \cdot m \cdot v^2\right) \\ &\approx 1.7\times 10^2\; \rm J \end{aligned}[/tex].
The sum of the translational and rotational kinetic energy of this cylinder would be approximately [tex]5.1\times 10^2\; \rm J[/tex].
