I began applying first law of thermodynamics (delQ=dE+delW) to a freely falling body. We see that for this case delQ=0, so dE=-delW. As dE consists of both potential and kinetic energies we see that neglecting any viscous losses, dE=0. (At max. height, KE=0,PE=mgh and at bottom most point PE=0,KE=.5*m*v^2=mgh). This concludes that delW=0. But what happens to the work done by gravity? Is there any restriction placed on delW to be used in first law? If you say that work done by the gravity is compensated by change in potential energy, how to convey this using first law of thermodynamics?