Calculating Minimum Motor Horsepower for Elevator Lifting

The question asks for the minimum horsepower required by a motor to lift an elevator weighing 500 kg at a constant velocity of 0.25 m/s, assuming no frictional losses and taking the acceleration due to gravity (G) as 9.8 m/s².

Understanding the Problem: Force and Power

When an object is lifted at a constant velocity, the net force acting on it is zero. This means the upward force applied by the motor must be equal in magnitude to the downward force due to gravity (the weight of the elevator).

The weight of the elevator is given by:

\( \text{Weight} = \text{mass} \times \text{acceleration due to gravity} \)

\( W = m \times g \)

The force required to lift the elevator at constant velocity is equal to its weight. So, the upward force \(F\) exerted by the motor is:

\( F = W = 500 \, \text{kg} \times 9.8 \, \text{m/s}^2 \)

\( F = 4900 \, \text{N} \)

Power is the rate at which work is done. When a constant force \(F\) acts on an object moving at a constant velocity \(v\) in the direction of the force, the power \(P\) is given by:

\( P = F \times v \)

Calculating Power in Watts

Using the force calculated and the given velocity, we can find the power required in Watts:

\( P = 4900 \, \text{N} \times 0.25 \, \text{m/s} \)

\( P = 1225 \, \text{Watts} \)

Converting Power from Watts to Horsepower

The power calculated is in Watts. We need to convert this to horsepower (hp). The standard conversion factor is approximately 1 horsepower ≈ 746 Watts.

To convert Watts to horsepower, we divide the power in Watts by the conversion factor:

\( \text{Power in hp} = \frac{\text{Power in Watts}}{\text{Conversion factor}} \)

\( \text{Power in hp} = \frac{1225 \, \text{W}}{746 \, \text{W/hp}} \)

\( \text{Power in hp} \approx 1.641 \, \text{hp} \)

The minimum horsepower of the motor to be used is approximately 1.64 hp. This value matches one of the given options.

Conclusion on Minimum Motor Horsepower

Based on the calculations, the minimum horsepower required for the motor to lift the elevator at a constant velocity of 0.25 m/s is approximately 1.64 hp.

Revision Table: Elevator Power Calculation

Additional Information: Work, Power, and Units

This problem involves the concepts of work and power. Work is done when a force causes displacement. In this case, the motor does work against gravity to lift the elevator.

• Work: Work (\(W\)) done by a constant force \(F\) acting over a displacement \(d\) in the direction of the force is \(W = F \times d\). The unit of work is the Joule (J).
• Power: Power (\(P\)) is the rate at which work is done, or the rate at which energy is transferred. \(P = \frac{\text{Work}}{\text{Time}}\). The unit of power is the Watt (W), which is equivalent to Joules per second (J/s).
• Power (Force and Velocity): As shown in the solution, if a force \(F\) acts on an object moving at velocity \(v\) in the direction of the force, power can also be calculated as \(P = F \times v\). This formula is very useful in problems involving lifting or pushing at a constant speed.
• Horsepower: Horsepower (hp) is a traditional unit of power, often used for engines and motors. One horsepower is approximately 746 Watts. This unit originated historically to compare the output of steam engines to the power of draft horses.

The problem assumes no frictional loss, which simplifies the calculation. In a real-world scenario, additional power would be needed to overcome friction in the elevator's moving parts and air resistance, meaning the actual required motor horsepower would be slightly higher than this minimum value.