Say you're pushing a heavy desk (225kg) across a carpeted floor. You either push it at 3m/s (say you applied a 675N force for 1sec, so you apply the force for 1.5m), or at 6 m/s (you applied 1350N force for 1sec, so the force is applied over 3m). After that one second impulse, however, you continue pushing it at a constant velocity (a total distance of 6m). After the initial second, are you just applying a force of FNuk to the object, or are you still pushing it at 675N or 1350N? I think that your initial 675N & 1350N forces are replaced by a new vector, equivalent but opposite, to the frictional force. How much work do you do on the object? I believe you would 1) multiply 675N x 1.5 m & 1350N x 3m; then 2) add FNuk x [6m - (1.5m || 3m)] (the work done after the initial acceleration) ...to get the total work you impart on the object in your two separate trials. However, due to friction, the force you apply is reduced, and the net K.E. for both objects is reduced by ukFN during part one. No work is done on the object during part two. This seems counter-intuitive, since you yourself are expending a lot of energy pushing an object a constant velocity, but no net work is being done on the object. Could someone confirm this for me? Thanks.