An electric winch is used to lift a cargo container of mass 120 kg vertically upwards at a constant speed of 1.5 m s-1. Given that the winch is 70% efficient, find the minimum electrical power input to the winch motor.
Assume the acceleration due to gravity is g=9.8 m s−2g = 9.8\text{ m s}^{-2}g=9.8 m s−2.
1.23 kW1.23\text{ kW}1.23 kW
1.76 kW1.76\text{ kW}1.76 kW
2.52 kW2.52\text{ kW}2.52 kW
1.12 kW1.12\text{ kW}1.12 kW