Consider the system shown in the figure (Figure 1) . Block A weighs 45.1 N and b
ID: 1371710 • Letter: C
Question
Consider the system shown in the figure (Figure 1) . Block A weighs 45.1 N and block B weighs 29.7 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 45.1 N , falls asleep on top of block A. If block B is now set into downward motion, what is its acceleration magnitude?
c)A cat, also of weight 45.1 N , falls asleep on top of block A. If block B is now set into downward motion, what is its acceleration direction? upward or downwards
Explanation / Answer
a) blocks are moving with constant speed that means a =0
on block B,
T - mBg = 0
T = mBg = 29.7 N
now on block A,
friction - T = 0
and friction = u N = u mA g
u mAg = T
u(45.1) = 29.7
u =0.659
b) now cat is sleeping on block A,
new N = 45.1 + 45.1 = 90.2 N
friction = uN = 0.659 x 90.2 = 59.4 N
now on A,
T - 59.4 = (90.2/9.81)a
on B,
29.7 - T = (29.7/9.81)a
12.22a = - 29.7
a = 2.43 m/s^2
c) direction - ? upwards