Does a force do work on an object with constant velocity? I know that a force does no work on an object if the object's displacement is zero, but if an object is moving at a constant velocity $\bar{v}$, and a force $\bar f$ (let's say that $\bar f$ and $\bar v$ have the same direction) acts upon it over a linear distance $d$, but the velocity remains the same (perhaps there is an equal and opposing force $\bar f_{opp}$), does $\bar f$ do work on the object?
 A: Yes, though in many cases the object will end up transferring the work to something else.
For example suppose you are pushing a boat through water with a force $F$ at a speed $v$. The the power (i.e. work per second) you are supplying to the boat is simply $Fv$. However the boat is in turn pushing on the water and doing work on the water. The work you put in gets transferred through the boat and ends up heating the water.
Response to comment:
Good question, and I think the answer is that no, this isn't work in the usual sense. Friction is basically an adhesive process. Down at the atomic scale the atoms in your box bond with the atoms of the substrate by the usual interatomic forces. To move the box you have to break these bonds and that dissipates energy.
Having said this, there is often a second component to friction i.e. viscous losses in the bulk. If you're pushing the box over a rubber surface the rubber will deform in response to the force exerted by the box, then snap back as the box moves away. In this case the does work on the subtrate is the usual way by deforming it.
A: The above answer is  correct , though you may like to hear what i am saying :
Suppose a person is applying force 5 Newton in -> direction. Friction is 5 Newton in <- direction.
(( The person had to start moving box by applying a slightly larger force: say 5.001 Newton at time=0.
so that an instantaneous acceleration increases the velocity to some constant value. ))
now lets say velocity is 1 m/s.
lets calculate Work done by "Friction" in 2 sec ( only by friction )
distance travelled = 1 x 2 = 2 m
Friction Force is opposite to displacement vector , so ,
Work done by friction = 5 x 2 x (cos(180)) = -10 J
Similarly work by person is = 10 J
You see work done by friction or work done by person is not zero.
BUT Net work done on that box is 0 ( sum of all work done by the forces)
