Question

Какой минимальный КПД должен иметь мотор, потребляющий электрическую мощность 2 кВт, для того, чтобы поднять груз массой 50 кг со скоростью 3 м/с?

Answer 1

Решение..................................................

Answer 2

Minimum Efficiency of Motor Required to Lift a 50kg Load at 3m/s with 2kW Power Consumption

To calculate the minimum efficiency of a motor required to lift a 50kg load at a speed of 3m/s with a power consumption of 2kW, we can use the formula for mechanical power:

Mechanical Power (Pm) = Force × Velocity

The force required to lift the load can be calculated using the formula:

Force (F) = Mass × Acceleration

Given: - Mass (m) = 50kg - Velocity (v) = 3m/s - Power Consumption (P) = 2kW

We can calculate the force required to lift the load and then determine the minimum efficiency of the motor.

Calculation

First, let's calculate the force required to lift the load: F = m × a

Using the formula for mechanical power: Pm = F × v

We can then calculate the minimum efficiency of the motor using the formula: Efficiency = (Mechanical Power Output / Electrical Power Input) × 100%

Result

The minimum efficiency of the motor required to lift a 50kg load at a speed of 3m/s with a power consumption of 2kW is calculated to be approximately 75%.

This calculation is based on the given parameters and the formulas for mechanical power and efficiency.