A lawnmower engine with an efficiency of 0.29 rejects 5500 J of heat every second. What is the magnitude of the work that the engine does in one second?

Answer 1

.29=(input-waste)/input
solving for input..

input(1-.29)=waste
but work=.29 input

work/.29 (.71)=5500

work=.29*5500/.71

check that