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. 。