[SOLVED] Work done on box 1. The problem statement, all variables and given/known data An 8.00kg package in a mail-sorting room slides 2.00m down a chute that is inclined at 53.0 below the horizontal. The coefficient of kinetic friction between the package and the chute's surface is 0.400. its asking me for the work done on the package by gravity and then the net work done on the package. wouldnt the work by gravity just be the weight. and for the normal force i just subtracted the frictional force from the normal force. where am i messing up? 2. Relevant equations W= F cos(53) s Friction Force= .4(Fnormal) 3. The attempt at a solution I already figured out the frictional force to be 37.7 and the force 94J.