Consider the system shown in ?Fig. E5.34.? Block A weighs 45.0 N, and block B weighs 25.0 N. Once block B is set into downward motion, it descends at a constant speed. (a) Calculate the coefficient of kinetic friction between block A and the tabletop. (b) A cat, also of weight 45.0 N, falls asleep on top of block A . If block B is now set into downward motion, what is its acceleration (magnitude and direction)?
Solution 34E Step 1: a) eight of block A = 45 N Tension on the rope due to block B = 25 N We know that, Force × coefficient of kinetic friction = Resultant tension on the rope That is, F× = T This is applicable only if, the other block is moving with constant speed. That is, no acceleration. Therefore, 45 N × = 25 N Rearranging the equation, = 25 N / 45 N = 5/9 = 0.544
