Physics · Work, Energy and Power
An elevator weighing 500 kg is to be lifted up at a constant velocity of 0.25 m/s. What would be the minimum horsepower of the motor to be used? (Take g = 9.8 m/s2 and assume there is no frictional loss, 1 hp = 746 W)
Verified Solution4 OptionsUpdated: October 2026👁️1 View
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- 12.00 hp
- 21.92 hp
- 31.64 hp✓
- 41.00 hp
Correct answer
1.64 hp
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