NVAEasyJEE Main 2023 · 10 April, Shift 2Power

Physics Question from JEE Main 2023 · 10 April, Shift 2

If the maximum load carried by an elevator is 1400kg1400 \, \text{kg} (600kg600 \, \text{kg} - Passengers + 800kg800 \, \text{kg} - elevator), which is moving up with a uniform speed of 3m s13 \, \text{m s}^{-1} and the frictional force acting on it is 2000N2000 \, \text{N}, then the maximum power used by the motor is _____ kW\text{kW} (g=10m/s2g = 10 \, \text{m/s}^2).

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