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(D) For a body rolling without slipping,the total kinetic energy is $K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})

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$\frac{14}{15}$

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(D) For a body rolling without slipping,the total kinetic energy is $K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$,where $k$ is the radius of gyration.By conservation of energy,$mgh = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$,so $h = \frac{v^2}{2g}(1 + \frac{k^2}{R^2})$.For a solid sphere,$I = \frac{2}{5}mR^2$,so $k^2 = \frac{2}{5}R^2$. Thus,$h_{sph} = \frac{v^2}{2g}(1 + \frac{2}{5}) = \frac{v^2}{2g}(\frac{7}{5})$.For a solid cylinder,$I = \frac{1}{2}mR^2$,so $k^2 = \frac{1}{2}R^2$. Thus,$h_{cyl} = \frac{v^2}{2g}(1 + \frac{1}{2}) = \frac{v^2}{2g}(\frac{3}{2})$.Therefore,the ratio is $\frac{h_{sph}}{h_{cyl}} = \frac{7/5}{3/2} = \frac{7}{5} 	imes \frac{2}{3} = \frac{14}{15}$.

$A$ solid sphere and a solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping throughout the motion. The two climb maximum heights $h_{sph}$ and $h_{cyl}$ on the incline. The ratio $\frac{h_{sph}}{h_{cyl}}$ is given by

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