A belt drive transmits 56kW from the driver pulley to the driven pulley. If 2.5kW is lost in belt friction, calculate:
1. kW's supplied by driver pulley
2. efficiency of the belt drive expressed as a percentage.
%Efficiency = (power output/power input) x 100%
The Attempt at a Solution
I'm really not sure how to tackle this question, if anyone can help it would be greatly appreciated.