When the voltage of 0.3V is applied, the current is 0.3A, and the motor does not rotate. When the voltage of 2.0V is applied at both ends of the motor, the current is 0.8A, and the motor works normally. What is the efficiency of the vacuum cleaner?

When the voltage of 0.3V is applied, the current is 0.3A, and the motor does not rotate. When the voltage of 2.0V is applied at both ends of the motor, the current is 0.8A, and the motor works normally. What is the efficiency of the vacuum cleaner?

When the motor does not run, according to Ohm's law, the resistance of the motor is r = u1i1 = 1 Ω. When the motor works normally, the output power of the motor is p output = ui-i2r = 2 × 0.8-0.82 × 1 = 0.96 (W), and the total power of the motor is p total = UI = 1.6W, so the efficiency of the vacuum cleaner is η = P output P total ×% = 60%. A: the efficiency of the vacuum cleaner is 60%

What is the external work of motor
I'm a bit stupid,

Well, sometimes physics is easy to let us into a dead end
My understanding is as follows:
First of all, we know that the motor is a device that converts electrical energy into mechanical energy. We are not very clear about its structure, but we can be sure that there must be loss in the conversion process. The loss lies in the internal resistance of the motor (just look at it as a resistance as a whole)
In the circuit, part of the electric energy consumed by the motor is used for resistance heating, and the other part is used to convert it into mechanical energy
In other words: the external work of the motor refers to the part of the motor that converts electrical energy into mechanical energy
In the process of solution, the total electric energy consumption (voltage * current * time) of the branch where the motor is located is subtracted from the heat generated by the internal resistance of the motor (square of current * internal resistance * time)
I don't know if this is what the landlord asked