Work Done By Gravitational Force and Terminal Velocity Using these two definitions of work:

*

*Work = Force * Distance


*Work = Change in a particle's kinetic energy
When a skydiver hits terminal velocity, their change in kinetic energy is basically fixed for the rest of their path after that point (before they land). So using the second definition, it would seem that the work done by gravitational force is constant for some time. But according to the first definition, the work done by gravitational force should be increasing because the skydiver's displacement through the air is increasing.
How do we reconcile this? What is the actual work done by gravitational force?
 A: Your first equation is the definition of the work done by a force; the second one is the result of the Work-Energy Theorem, which relates the net work done on an object to its change in kinetic energy. Because the Work-Energy Theorem uses the net work (i.e. from all forces) done on an object, it cannot be used to find the work done by a single force, unless the work done by all other forces is known.
In your example of a skydiver falling at terminal velocity, the work done by gravity increases with time, but the net work done on the skydiver remains constant, because drag does negative work equal in magnitude to gravity.
A: The work done by gravity is still the force times the distance, or $mgh$ (where $m$ is the mass of the person and $h$ is the height from which they fell). However, the reason terminal velocity exists is because there's air resistance to counteract gravity. When the skydiver reaches terminal velocity, the air resistance is so great that it completely cancels out the gravitational force, hence producing 0 net acceleration. Here, gravity is still doing positive work, but air resistance (friction) is doing negative work. That's why the net work - the change in kinetic energy - is less than the work done by gravity.
