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(C) When an object slides down an inclined plane without friction,its potential energy is converted entirely into translational kinetic energy: $mgh =

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The rectangular block will reach the bottom first.

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(C) When an object slides down an inclined plane without friction,its potential energy is converted entirely into translational kinetic energy: $mgh = \frac{1}{2}mv^2$,which gives $v_{slide} = \sqrt{2gh}$.When a solid sphere rolls down without slipping,its potential energy is converted into both translational and rotational kinetic energy: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.For a solid sphere,$I = \frac{2}{5}mr^2$ and $\omega = \frac{v}{r}$. Substituting these,we get $mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$.Solving for velocity,$v_{roll} = \sqrt{\frac{10}{7}gh} \approx \sqrt{1.43gh}$.Since $\sqrt{2gh} > \sqrt{1.43gh}$,the sliding block has a higher velocity throughout the motion and will reach the bottom first.

$A$ solid sphere rolls down from rest on an inclined plane. At the same time,a rectangular block slides down from rest on the same inclined plane. Which of the following is true?

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