Consider the system shown in the figure (Figure 1) . Block A weighs 46.1 N and block B weighs 28.7 N . Once block B is set into downward motion, it descends at a constant speed.
Calculate the coefficient of kinetic friction between block A and the tabletop
A cat, also of weight 46.1 N , falls asleep on top of block A. If block B is now set into downward motion, what is its acceleration magnitude
The Attempt at a Solution
∑F in x direction of A: -F+T = ma
-μ92.1+T = 92.2/ 9.8 a
∑F in y of A N-w = ma
92.2-9.4(9.8) = 0
∑F in Y of B T-w
T-2.93 (9.8) = 2.93 a
plug T into first equation: -.623(92.1)+28.7+2.93a = 9.4a
-.623(92.1)+28.7= (9.4-2.93) a
divide both sides
( -.623(92.1)+28.7) /9.4-2.93 = a
gives me wrong answer, where did I go wrong?