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Physics · WORK, ENERGY AND POWER
हिन्दीA loaded elevator with total mass \( 2000\text{ kg} \) moves upward at a constant speed of \( 3\text{ m s}^{-1} \). A constant frictional force of \( 5000\text{ N} \) opposes its motion. Taking \( g = 10\text{ m s}^{-2} \), the minimum power delivered by the motor in horse-power (hp) is approximately: (Take \( 1\text{ hp} = 746\text{ W} \))
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Correct answer · Option A
\( 100.5\text{ hp} \)
Explanation
Total downward force \( F = mg + f = 2000 \times 10 + 5000 = 25000\text{ N} \). Power \( P = F \cdot v = 25000 \times 3 = 75000\text{ W} \). In horsepower: \( P = \frac{75000}{746} \approx 100.5\text{ hp} \).
