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Determine the decrease in potential energy and the corresponding increase in kinetic energy for an object descending an inclined plane.

1. K.E. = 1/2 Iω² + 1/2 mv²
2. K.E. = 1/2 (1/2 mR²)(ω²) + 1/2 mv²
3. 1/4 mv² + 1/2 mv² = 3/4 mv²
4. 3/4 mv² = mgh, hence v = √(4/3)gh

Correct answer: 3/4 mv² = mgh, hence v = √(4/3)gh

Solution

The rolling object converts its potential energy (mgh) into both translational and rotational kinetic energy. The correct relation is given in option D, where the total kinetic energy equals 3/4 mv², and the velocity is derived as v = √(4/3)gh.

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