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(A) For an object rolling without slipping down an inclined plane of height $h$,the conservation of mechanical energy states that the potential energy

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$\sqrt{\frac{10}{7}gh}$

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(A) For an object rolling without slipping down an inclined plane of height $h$,the conservation of mechanical energy states that the potential energy at the top equals the sum of translational and rotational kinetic energy at the bottom.$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$Since the object is a sphere,its moment of inertia $I = \frac{2}{5}mR^2$ and the rolling condition is $\omega = \frac{v}{R}$.Substituting these into the energy equation:$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2$$mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2$$mgh = \frac{7}{10}mv^2$$v^2 = \frac{10}{7}gh$$v = \sqrt{\frac{10}{7}gh}$

What is the velocity of a sphere starting from rest and rolling without slipping down an inclined plane of vertical height $h$?

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