Question

A battery of ? = 2.10 V and internal resistance R = 0.600 ? is driving a motor. The motor is lifting a 2.0 N mass at constant speed v = 0.50 m/s. Assuming no energy losses, find the current i in the circuit.

(a) Enter the lower current.

(b) Enter the higher current.

(c) Find the potential difference V across the terminals of the motor for the lower current.

(d) Find the potential difference V across the terminals of the motor for the higher current.

Answer 1

Mass is not in Newtons, I'll assume you mean force of 2 Newtons.

In one second, the motor lifts with a force of 2 newtons over a distance of 0.5 meter, which is work/energy of Fd = 1 Joule.
Since this occurs every second, this is 1 Joule/second or 1 Watt of power.

Now we have a circuit with 2.10 volts in series with 0.6 ohms and a R dissipating 1 watt.

R total = R +0.6
I total = 2.1/(R +0.6)
Voltage across R=
VL = 2.1R/(R +0.6)

power = 2.1/(R +0.6) x 2.1R/(R +0.6) = 1
2.1