Assume that it is a simple dc motor.
The instant you switch the motor on a current $I$ will flow through the coil and a torque will be produced on the coil.
If the supply voltage is $V$ and the resistance of the coil is $R$ then the electrical power supplied by the supply is $VI$ and this is dissipated as heat in the coil if the motor $I^2R$.
If the coil was unable to move then this would be bad news for the motor as the heat generated might well be sufficient to melt the coil.
Once the coil starts to move then the voltage of the supply minus the (induced/back) emf produced by the coil rotating in a magnetic field $\mathcal E$ is equal to the voltage across the coil resistance $V -\mathcal E = IR$.
As the motor speed $\omega$ increases the induced emf increases because the $\mathcal E \propto \omega$.
So in terms of power $VI = \mathcal E I + I^2R$.
$VI$ is the power supplied the voltage source, $I^2R$ is the power dissipated as heat in the coil and $\mathcal E$ is the power for the motor to do work eg useful power lift your mass and wasted power overcoming friction at the baring.
You will note that the current through the rotating coil is now lower than than when the coil was not moving.
If you want the motor to do more work per second then the speed of rotation $\omega$ of the coil decreases and so the induced emf $\mathcal E$ decreases.
However the important thing to note is that the fractional increase in current $I$ is greater than the fractional decrease in the emf $\mathcal E$ so the product $\mathcal E I$ (the power output of the motor) increases.
If you have ever used an electric drill you perhaps have noticed that when the drill is under load (eg drilling a hole) it is rotating slower than when the drill is under no load conditions.
