Work--uniform speed? A 40.0 kg box is pushed 3.00m at uniform speed across a horizontal garage floor. and its then lifted 1 m into the back of a truck. Assuming that the force of friction acting between the box and the floor is 80.0 N what is the total work done in moving the box? F = ma W= Fd Ffriction= Ufnet (u= coiffencet of friction) So basically I started off by F = ma in which case accelration is = to zero because its going at a uniform speed now if thats the case then W = O(F)3(d) so that means work is zero as well. but then i was like how does friction work into that? i suppose its just to confuse me. Becuase what ever number i get at the begining i'd add to the W = MGH w = 40 x 9.81 x1 = 392 J but if it the answer to the first part is zero wouldn't the total work be just the 392 from lifting it to the truck?? Hopefully that made sense and yu can help me clear this up thanks!