A force F = (3.00 N)i + (7.00 N)j + (7.00 N)k acts on a 2.00 kg mobile object that moves from an initial position of di = (3.00 m)i – (2.00 m)i + (5.00 m)k to a final position of df = – (5.00 m)i + (4.00 m)i + (7.00 m)k in 4.00 s. Find
(a) The work done on the object by the force in the 4.00 s interval,
(b) The average power due to the force during that interval, and
(c) The angle between vectors di, and, df.