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Difficulty: MediumWork, Energy, and Power

A submersible drainage pump is used to clear water from a flooded basement. The pump lifts 200 kg200\text{ kg} of water to a street-level drain located 5 meters5\text{ meters} above the basement floor. If the pump operates at a constant useful power output of 500 W500\text{ W} and the acceleration due to gravity is 10 m/s210\text{ m/s}^2, how many seconds does it take to complete the lift?

  1. A
    2 seconds2\text{ seconds}
  2. B
    4 seconds4\text{ seconds}
  3. 20 seconds20\text{ seconds}Answer
  4. D
    200 seconds200\text{ seconds}

Answer

20 seconds20\text{ seconds}
Lifting an object at constant speed requires an upward force equal to its weight: F=mg=200 kg×10 m/s2=2000 NF = mg = 200\text{ kg} \times 10\text{ m/s}^2 = 2{}000\text{ N}. The work performed over a vertical height of 5 m5\text{ m} is W=Fd=2000 N×5 m=10000 JW = Fd = 2{}000\text{ N} \times 5\text{ m} = 10{}000\text{ J}. Since power represents the rate of work (P=WtP = \frac{W}{t}), the time required is t=WP=10000 J500 W=20 secondst = \frac{W}{P} = \frac{10{}000\text{ J}}{500\text{ W}} = 20\text{ seconds}.

Step-by-Step Solution

1
Calculate the downward gravitational force (weight) acting on the water
F=m⋅g=200 kg×10 m/s2=2000 NF = m \cdot g = 200\text{ kg} \times 10\text{ m/s}^2 = 2{}000\text{ N}
Lifting the water requires applying an upward force equal to its weight.
2
Calculate the total work done (gain in gravitational potential energy)
W=F⋅d=2000 N×5 m=10000 JW = F \cdot d = 2{}000\text{ N} \times 5\text{ m} = 10{}000\text{ J}
Work is the product of the upward force and the vertical displacement.
3
Calculate the time required using the definition of power (P=WtP = \frac{W}{t})
t=WP=10000 J500 W=20 secondst = \frac{W}{P} = \frac{10{}000\text{ J}}{500\text{ W}} = 20\text{ seconds}
Power is the rate at which work is performed over time.

Key Concept

Work is calculated as force times distance (W=Fd=mghW = Fd = mgh), and power is the rate of doing work (P=WtP = \frac{W}{t}), which allows solving for time as t=WPt = \frac{W}{P}.

Alternative Method

Use the gravitational potential energy equation directly: ΔPE=mgh=200 kg×10 m/s2×5 m=10000 J\Delta PE = mgh = 200\text{ kg} \times 10\text{ m/s}^2 \times 5\text{ m} = 10{}000\text{ J}. Then apply t=ΔPEP=10000 J500 W=20 secondst = \frac{\Delta PE}{P} = \frac{10{}000\text{ J}}{500\text{ W}} = 20\text{ seconds}.
Estimated Time:1m 0s
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Topics

Work, Energy, and Power

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