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An electric motor has a rating of 4.0 × 10^2 watts. How much time will it take for this motor to lift a 50.-kilogram mass a vertical distance of 8.0 meters?
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hint: power \(W\) is the work done by an external force over time, so we can write: \[\Large W = \frac{L}{{\Delta t}}\] now, the work done by the motor, is equal to the work done by gravity, so we can write: \[\Large L = mgh\] where \(h=8\) is the vertical displacement. Therefore after a substitution, we get: \[\Large W = \frac{L}{{\Delta t}} = \frac{{mgh}}{{\Delta t}}\] and finally: \[\Large \Delta t = \frac{{mgh}}{W} = \frac{{5 \cdot 9.81 \cdot 8}}{400} = ...\sec \]
Remember units : Watt is J/s , also you get seconds in answer. Unit of mgh is J (joule)
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