A 1000 kg elevator rises from rest in the basement to the fourth floor, a distance of 20 m. As it passes the fourth floor its speed is 4 m/sec. There is a constant frictional force of 600 N. Then   percentage  of total work done is lost due to friction nearly is

total work done = work done against gravity+work done against friction + increase in kinetic energy                         =mgh+Ffriction×h +12mv2 =1000×9.8×20+600×20 +121000×16     =1039.8×20 + 12 +8  =103×216     % of work done against friction   12216×100=5.5