Given F=(2ti^+6t2j^) N and m=2 kg.
Acceleration of the body is a=mF=22ti^+6t2j^=ti^+3t2j^ m/s2.
Since the body begins to move from rest, its velocity at time t is given by v=∫adt=∫(ti^+3t2j^)dt=2t2i^+t3j^ m/s.
At t=2 s, the force is F=2(2)i^+6(2)2j^=4i^+24j^ N.
At t=2 s, the velocity is v=2(2)2i^+(2)3j^=2i^+8j^ m/s.
Power produced by the force is P=F⋅v=(4i^+24j^)⋅(2i^+8j^).
P=(4×2)+(24×8)=8+192=200 W.
Answer: 200