(a) Find the useful power output of an elevator motor that lifts a 2500 kg load a height of 45.0 m in 12.0 s, if it also increases the speed from rest to 4.00 m/s. Note that the total mass of the counterbalanced system is 10000 kg--so that only 2500 kg is raised in height, but the full 10000 kg is accelerated.