Question

The elevator of a large building is to raise a net mass of \(400 \mathrm{kg}\) at a constant speed of \(12 \mathrm{m} / \mathrm{s}\) using an electric motor. Minimum power rating of the motor should be \((a) 0 \mathrm{kW}\) \((b) 4.8 \mathrm{kW}\) \((c) 47 \mathrm{kW}\) \((d) 12 \mathrm{kW}\) \((e) 36 \mathrm{kW}\)

Short answer

Answer: (c) 47 kW

Step-by-step solution

Calculate the gravitational force on the elevator

First, we need to find the gravitational force acting on the elevator. We can do this using the formula \(F=mg\), where \(F\) is the gravitational force, \(m\) is the mass of the elevator, and \(g\) is the acceleration due to gravity, which is \(9.81 \mathrm{m/s^2}\). With the given mass of \(400 \mathrm{kg}\), the gravitational force can be calculated as: \(F= (400 \mathrm{kg})(9.81 \mathrm{m/s^2})\) \(F= 3924 \mathrm{N}\)

Calculate the minimum power needed

Now that we have the gravitational force, we can calculate the minimum power needed to lift the elevator at a constant speed using the formula \(P=Fv\). The given speed of the elevator is \(12 \mathrm{m/s}\). So, the power required is: \(P= (3924 \mathrm{N})(12 \mathrm{m/s})\) \(P= 47088 \mathrm{W}\)

Convert power to kilowatts

Since the answer choices are given in kilowatts, we need to convert the calculated power from watts to kilowatts. There are \(1000\) watts in a kilowatt. To convert, simply divide the power in watts by \(1000\): \(P= \dfrac{47088 \mathrm{W}}{1000}\) \(P= 47.088 \mathrm{kW}\)

Choose the correct answer

Since we found that the minimum power needed is \(47.088 \mathrm{kW}\), we can choose the closest answer to our calculated value: Choice (c) \(47 \mathrm{kW}\) is the correct answer.