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Zorluk: OrtaWork, Energy, and Power

An electric winch with a power rating of 400 W400\text{ W} is used to lift a 120 kg120\text{ kg} engine block vertically by a distance of 5 meters5\text{ meters} at a constant speed. Assuming the acceleration due to gravity is 10 m/s210\text{ m/s}^2, how much time does it take to complete the lift?

  1. A
    1.5 seconds1.5\text{ seconds}
  2. B
    7.5 seconds7.5\text{ seconds}
  3. 15 seconds15\text{ seconds}Cevap
  4. D
    24 seconds24\text{ seconds}

Cevap

15 seconds15\text{ seconds}
The weight of the 120 kg120\text{ kg} engine block is F=mg=120 kg×10 m/s2=1200 NF = mg = 120\text{ kg} \times 10\text{ m/s}^2 = 1{}200\text{ N}. Lifting it 5 meters5\text{ meters} requires W=Fd=1200 N×5 m=6000 JW = Fd = 1{}200\text{ N} \times 5\text{ m} = 6{}000\text{ J} of work. Since power is the rate of work (P=WtP = \frac{W}{t}), the time required is t=WP=6000 J400 W=15 secondst = \frac{W}{P} = \frac{6{}000\text{ J}}{400\text{ W}} = 15\text{ seconds}.

Adım Adım Çözüm

1
Calculate the downward gravitational force (weight) of the engine block
F=m×g=120 kg×10 m/s2=1200 NF = m \times g = 120\text{ kg} \times 10\text{ m/s}^2 = 1{}200\text{ N}
Lifting the object at constant speed requires an upward force equal to its weight.
2
Calculate the total mechanical work performed to lift the engine
W=F×d=1200 N×5 m=6000 JW = F \times d = 1{}200\text{ N} \times 5\text{ m} = 6{}000\text{ J}
Work is defined as force multiplied by displacement in the direction of the force.
3
Calculate the time required using the definition of power
t=WP=6000 J400 W=15 secondst = \frac{W}{P} = \frac{6{}000\text{ J}}{400\text{ W}} = 15\text{ seconds}
Power is the rate of doing work (P=WtP = \frac{W}{t}), so time equals work divided by power.

Anahtar Kavram

Power as the rate of doing work (P=WtP = \frac{W}{t}) combined with gravitational potential work (W=mghW = mgh)
Tahmini Süre:1m 0s
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Konular

Work, Energy, and Power

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