Block A has weight w_A and block B has weight w_B. Block A is positioned on the horizontal surface of a table. Block A is connected to a cord passing over an easily turned pulley that has Block B hanging down from the pulley. Once Block B is now set into downward motion, it descends at a constant speed. a. Calculate the coefficient of kinetic friction between Block A and the tabletop. I drew Block A's force diagram. For this part, I am assuming Block A is not in motion as yet; it is just sitting on the tabletop. (Is this correct?) Well, F_fr = mu_K * F_N and the normal force = w_A. F_fr = mu_k*w_A mu_k = F_fr/w_A But I think this is somewhat incorrect. Musn't I express mu_k with the known variables give. What is the Value for F_fr? b. A cat, also of weight w_A, falls asleep on top of block A. If Block B is now set into downward motion, what is its acceleration (magnitude and direction)? Express a in w_A, w_B, and g. Well, I am unsure of the F_fr expression. This is what I did for this part. For Block A, the expression is now F_T - F_fr = [(2*w_A)/g]*a For Block B, ---------------------w_B - F_T = (w_B/g)*a Adding these two expressions, I get w_B - F_fr = [(2*w_A)/g]*a + (w_B/g)*a w_B - F_fr = a[(2*w_A + w_B)/g] a = [w_B - F_fr]/[(2*w_A + w_B)/g] a = g*[(w_B-F_fr)/(2*w_A + w_B)] But I don't know what to put for F_fr as noted in the first part. Are my reasonings and math that I did do correct? Thanks.