An electric motor lifts a crate with a weight of 400 N vertically upward through a height of 5.0 meters. If this operation takes exactly 4.0 seconds, what is the useful mechanical power output of the motor?

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• A 320 W
• B 500 W
• C 2000 W
• D 8000 W

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Answer: B. 500 W

Work done = force × distance = 400 N × 5.0 m = 2000 J. Power = work ÷ time = 2000 J ÷ 4.0 s = 500 W.

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