Yes.At 60% speed,the power required by load is proportional to cube of o.6 i.e 0.22 of that at rated speed only.Let us take an example.Let KW of load be 10 kW at rated speed.If its speed reduces to 60%,its power requirement is 2.2kW only.Let it be driven by a 10kwmotor.The current corresponding to 2.2 kW can be calculated taking into account the reduced efficiency and power factor of the motor for this reduced load by using the formula for input power to motor.Can you demonstrate any suitable values for this reduced efficiency and power factor to be used in that formula?
This is how I do it:

It's actually from a workbook with three tabs and that's the results part of one of them.
The Tqerr button (top right) runs a Visual Basic routine to get the slip right to match the load torque.
The data is for a real motor that has now been in service for quite a few years in a water pumping application.
The purpose of the calculations was to provide guaranteed efficiencies at various operating points at the bid stage of the project. Getting them wrong isn't an option.