A force F = (4.51 +6.1) + 5.0) (4.5+ 6.1 + 5.0) N acts on a 3.1kg object +5.0k N acts on a 3.1kg object that moves in 2.6s from an initial position 7 = (3.71 + 1.6) + 3.5k) m to a final position 2 = (4.31 +4.8 + 7.4) m. Find (a) the work done on the object by the force in that time interval, (b) the average power due to the force during that time interval, and (c) the angle between the vectors and 2. (a) Number Units (b) Number i Units (c) Number i Units >