flcosθ-μ(mg+fsinθ)L = 12mv2B

From Newton's second law at point B.

f？ n？ mg=mv2BR？

Simultaneous solution FN=35.6N n n.

According to Newton's third law, the pressure on a circular orbit is 35.6 N;

(2) In the process from B to C, it is obtained by kinetic energy theorem.

-mgR-Wf=0- 12mv2B

Substitute the data to get WF =1.4j;

Answer: (1) When the object reaches point B, the pressure on the circular arc orbit is 35.6 N. 。

(2) The work done by an object to overcome the frictional resistance on the arc trajectory BC is1.4j. 。