If the engine power is 3.3kW and it is 60% efficient, how much water will it pump in 5s from a height of 10m?
Correct answer: B. 100kg
- A. 60kg
- B. 100kg
- C. 75kg
- D. 80kg
Explanation
To calculate the amount of water pumped in 5 seconds from a height of 10 meters, we can use the principle of work and energy. First, let's determine the potential energy of the water at a height of 10 meters. Potential energy (PE) = mass (m) * gravitational acceleration (g) * height (h) where: g ≈ 9.81 m/s² (gravitational acceleration) Given that the potential energy is transferred into the water by the engine, and the engine is 60% efficient, we can calculate the work done by the engine (W_engine) as follows: W_engine = PEPE/efficiency , work done by the engine is also equal to the power (P) of the engine multiplied by time (t): W_engine = P * t We are given that the engine power (P) is 3.3 kW, and the time (t) is 5 seconds. Now we can set up the equation: 3.3 kW * 5 s = PE / 0.60 Solving for PE: PE = 3.3 kW * 5 s * 0.60 ≈ 9.9 kJ Now, we can find the mass of the water (m) using the potential energy (PE): PE = m * g * h 9.9 kJ = m * 9.81 m/s² * 10 m Solving for m: m = 9.9 kJ / (9.81 m/s² * 10 m) ≈ 0.101 kg ≈ 100 g
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Work transfers energy when a force causes displacement, while kinetic energy, gravitational potential energy and power describe motion, position and the rate of energy transfer. The work-energy theorem links net work with change in kinetic energy, and efficiency accounts for energy losses rather than treating them as energy destruction.
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