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(A) To climb the inclined surface to a height $h$,the total initial kinetic energy of the rolling sphere must be at least equal to the potential energ

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

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(A) To climb the inclined surface to a height $h$,the total initial kinetic energy of the rolling sphere must be at least equal to the potential energy gained at height $h$.The total kinetic energy of a rolling body is the sum of its translational and rotational kinetic energies:$K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$For a solid sphere,the moment of inertia $I = \frac{2}{5}mr^2$ and for pure rolling,$\omega = \frac{v}{r}$.Substituting these values:$K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2$$K_{total} = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$By the law of conservation of energy,the total kinetic energy must be greater than or equal to the potential energy $mgh$ at the top:$\frac{7}{10}mv^2 \ge mgh$$v^2 \ge \frac{10}{7}gh$$v \ge \sqrt{\frac{10}{7}gh}$

$A$ solid sphere is rolling on a frictionless surface,as shown in the figure,with a translational velocity $v \, m/s$. If it is to climb the inclined surface to a height $h$,then $v$ should be:

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