The figure here shows an overhead view of three horizontal forces acting on a cargo canister that was initially stationary but that now moves across a frictionless floor. The force magnitudes are F1 = 3.00 N, F2 = 3.60 N, and F3 = 10.0 N, and the indicated angles are 02 = 51.0 and 03 = 33.0. What is the net work done on the canister by the three forces during the first 3.50 m of displacement? Fo Number i ?? ? Unit J e Textbook and Media Assistance Used Hint Assistance Used The net work is the sum of the individual works (the work done by each force during the displacement).Key Ideas . Work W is energy transferred to or from an object via a force acting on the object. Energy transferred to the object is positive work, and from the object, negative work. . The work done on a particle by a constant force F during displacement d is W = Fd cos $ = F . d (work, constant force), in which o is the constant angle between the directions of F and d. . Only the component of F that is along the displacement d can do work on the object. . When two or more forces act on an object, their net work is the sum of the individual works done by the forces, which is also equal to the work that would be done on the object by the net force F net of those forces. For a particle, a change AK in the kinetic energy equals the net work W done on the particle: AK = Kf - Ki = W (work-kinetic energy theorem ) , in which K; is the initial kinetic energy of the particle and Kris the kinetic energy after the work is done. The equation rearranged gives us Kf = Ki + W