Given that,
Mass of object = 9.90 kg
Time =5.40 s
Suppose the force is (2.00i + 9.00j + 5.30k) N, initial position is (2.70i - 2.90j + 5.50k) m and final position is (-4.10i + 3.30j + 5.40k) m.
We need to calculate the displacement
Using formula of displacement

Where,
= initial position
= final position
Put the value into the formula


(a). We need to calculate the work done on the object
Using formula of work done

Put the value into the formula



(b). We need to calculate the average power due to the force during that interval
Using formula of power

Where, P = power
W = work
t = time
Put the value into the formula


(c). We need to calculate the angle between vectors
Using formula of angle

Put the value into the formula


Hence, (a). The work done on the object by the force in the 5.40 s interval is 41.67 J.
(b). The average power due to the force during that interval is 7.71 Watt.
(c). The angle between vectors is 79.7°