An athlete lifts a 100-kg barbell vertically 0.8 m in 0.5 seconds. Using g = 9.81 m/s², what is the approximate average mechanical power output?
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Answer: D
About 160 W: This calculates 100 kg × 0.8 m ÷ 0.5 s using mass instead of force (weight). Force = mass × g = 981 N.
About 392 W: This multiplies work by time (784.8 J × 0.5 s) instead of dividing. Power = work ÷ time.
About 785 W: 784.8 J is the work done, not power. It still needs to be divided by the 0.5 s duration.
About 1,570 W: Correct. Force = 100 × 9.81 = 981 N; work = 981 N × 0.8 m = 784.8 J; power = 784.8 J ÷ 0.5 s ≈ 1,570 W.
Work (J) = force (N) × displacement (m); power (W) = work ÷ time. Convert mass to force with g = 9.81 m/s².