A 68-kg skydiver falls at a constant terminal speed of 59 m/s. At what rate is the internal energy of the skydiver and surrounding air increasing?

The rate of change of internal energy of the skydiver is equal to the rate of change of gravitational potential energy of the skydiver.
To find out the rate of change of internal energy (dE/dt) of the skydiver, substitute 68-kg for mass of the skydiver m , 9.8 m/s² for the acceleration due to gravity g and 59 m/s for the terminal speed v in the equation,
dE/dt = mgv

Rounding off to two significant figures, the change of internal energy of the skydiver will be 39000 J/s.