Final answer:
The net work done on the elevator during its trip from the first floor to the fourth floor, assuming no energy losses, is the work required to overcome gravitational force for the vertical distance traveled, which is w = mgh.
Step-by-step explanation:
To determine the net work done on an elevator of mass m while it moves upward through a distance of three floors (each floor being h units apart), we can use the concept of work-energy principle. The work done by gravity will be negative, as it acts in the opposite direction of the displacement. However, the work done by the elevator system's motor to overcome this force will be equal in magnitude and opposite in sign, resulting in a network of zero, assuming no energy losses due to friction or air resistance.
During the ascent, the elevator accelerates from rest to a constant velocity, which requires work to be done to give it kinetic energy. However, once at a constant velocity, no network is done because the upward force by the elevator cable is equal to the weight of the elevator. When the elevator comes to rest on the fourth floor, the motor works to decelerate the elevator to zero velocity, converting its kinetic energy back into potential energy. Since the initial and final kinetic energies are zero (the elevator starts and stops at rest), the net work done is just the work needed to raise the elevator against gravity, minus the work done by gravity during the descent.
If we assume that the net work done by non-conservative forces (like friction if any) is negligible, the net work done on the elevator by the elevator system while going up would simply be the work done against gravity to move the elevator from the first to the fourth floor, which is w = mg, where g is the acceleration due to gravity.