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Zorluk: OrtaWork, Energy, and Power

Four electric hoists in a maintenance facility lift cargo vertically at constant speeds. Arrange the hoists in order of their average mechanical power output, from lowest power to highest power.

  1. 1A hoist exerting a force of 500 N500\text{ N} to lift a crate 4 m4\text{ m} in 5 s5\text{ s}
  2. 2A hoist exerting a force of 250 N250\text{ N} to lift a crate 10 m10\text{ m} in 5 s5\text{ s}
  3. 3A hoist exerting a force of 300 N300\text{ N} to lift a crate 6 m6\text{ m} in 3 s3\text{ s}
  4. 4A hoist exerting a force of 800 N800\text{ N} to lift a crate 3 m3\text{ m} in 3 s3\text{ s}

Cevap

The correct sequence from lowest to highest mechanical power is the hoist delivering 400 W400\text{ W} (500 N×4 m/5 s500\text{ N} \times 4\text{ m} / 5\text{ s}), followed by the hoist delivering 500 W500\text{ W} (250 N×10 m/5 s250\text{ N} \times 10\text{ m} / 5\text{ s}), then the hoist delivering 600 W600\text{ W} (300 N×6 m/3 s300\text{ N} \times 6\text{ m} / 3\text{ s}), and ending with the hoist delivering 800 W800\text{ W} (800 N×3 m/3 s800\text{ N} \times 3\text{ m} / 3\text{ s}).
Mechanical power equals the rate of doing work (P=F×dtP = \frac{F \times d}{t}). Computing the power output yields 400 W400\text{ W} for the 500 N500\text{ N} load over 4 m4\text{ m} in 5 s5\text{ s}, 500 W500\text{ W} for the 250 N250\text{ N} load over 10 m10\text{ m} in 5 s5\text{ s}, 600 W600\text{ W} for the 300 N300\text{ N} load over 6 m6\text{ m} in 3 s3\text{ s}, and 800 W800\text{ W} for the 800 N800\text{ N} load over 3 m3\text{ m} in 3 s3\text{ s}. This produces the ascending sequence 400 W→500 W→600 W→800 W400\text{ W} \rightarrow 500\text{ W} \rightarrow 600\text{ W} \rightarrow 800\text{ W}.

Adım Adım Çözüm

1
Recall the formula for mechanical power in terms of force, distance, and time.
P=Wt=F×dtP = \frac{W}{t} = \frac{F \times d}{t} where power (PP) is in Watts (W\text{W}), force (FF) in Newtons (N\text{N}), distance (dd) in meters (m\text{m}), and time (tt) in seconds (s\text{s}).
Power measures the rate at which work is performed.
2
Calculate the power output for each hoist.
First hoist: (300 N×6 m)/3 s=1800 J/3 s=600 W(300\text{ N} \times 6\text{ m}) / 3\text{ s} = 1800\text{ J} / 3\text{ s} = 600\text{ W}. Second hoist: (500 N×4 m)/5 s=2000 J/5 s=400 W(500\text{ N} \times 4\text{ m}) / 5\text{ s} = 2000\text{ J} / 5\text{ s} = 400\text{ W}. Third hoist: (800 N×3 m)/3 s=2400 J/3 s=800 W(800\text{ N} \times 3\text{ m}) / 3\text{ s} = 2400\text{ J} / 3\text{ s} = 800\text{ W}. Fourth hoist: (250 N×10 m)/5 s=2500 J/5 s=500 W(250\text{ N} \times 10\text{ m}) / 5\text{ s} = 2500\text{ J} / 5\text{ s} = 500\text{ W}.
Determines the exact numerical power for all four mechanical scenarios.
3
Arrange the resulting power values in ascending order.
400 W<500 W<600 W<800 W400\text{ W} < 500\text{ W} < 600\text{ W} < 800\text{ W}.
Satisfies the requirement to order from lowest power to highest power.

Anahtar Kavram

Power is the rate of doing work, defined as work divided by time (P=Wt=F×dtP = \frac{W}{t} = \frac{F \times d}{t}), and is measured in Watts.
Tahmini Süre:1m 15s
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