Since Work= F displacement cos theta. If a person runs up a flight of stairs and then back with a constant velocity, does that mean his acceleration is also 0. If that's the case is the person's force 0, making his work also 0. This doens't make sense to me. Or is acceleration, one due to gravity, so it would be 9.8 m/s^2? Does this change when a person goes down a flight of stairs or is the same?