(1) Current flowing through R I = ER+R = BLVR+R From the meaning of the question, if the current I flowing through R increases evenly with time, it can be known that the velocity V of the rod ab increases evenly with time and the acceleration of the rod is constant, so the metal rod does uniform acceleration motion.

(2) The magnitude of the ampere force on the rod is fa = bil = bblvr+rl = b2l2vr+r+R..

For the rod, according to Newton's second law, F+mgsinθ-FA=ma,

Substituting F=0.5v+2 gives: 2+mgsinθ+(0.5-B2L2R+r)v=ma,

Because the acceleration a of uniform acceleration motion has nothing to do with v,

So a=2+mgsinθm=8m/s2,

0.5-B2L2R+r=0，

R = 0.3 ω？

(3) From x= 12at2, the required time is t = 2xa = 2× 18s = 0.5s, and the speed of the ab rod is v = at = 4m/s..

The total joule heat in the circuit is q = r+rrq1= 0.3+0.20.3× 0.8j = 43j.

According to the law of conservation of energy: W+mgxsinθ=Q+ 12mv2?

Solution: w =103j ≈ 3.3j.

Answer: (1) The metal bar accelerates uniformly.

(2) The resistance value of resistor R is 0.3 Ω.

(3) The time t required for this process is 0.5s, and the work w done by pulling force F is 3.3J 。