Consider the system shown in the figure (Figure 1) . Block A weighs 43.6N and bl
ID: 1316959 • Letter: C
Question
Consider the system shown in the figure (Figure 1) . Block A weighs 43.6N and block B weighs 30.4N . 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 43.6N , 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 43.6N , falls asleep on top of block A. If block B is now set into downward motion, what is its acceleration direction?
Explanation / Answer
Once in motion, for block B to have a constant velocity, there must be no net force acting on it.
So FfA = WB (FfA is force of kinetic friction of block A: WB is weight of block B)
FfA = WB
mu sub k x Normal Force A = WB
mu sub k = WB / NA = 30.4 N /43.6 N = .6972477064
Cat situation...
FfA now equals:
FfA = mu sub k x (WA + WC) (WA is weight of block A: WC is weight of cat)
FfA = .6972477064 x (43.6 N x 2) = 60.8 N
FnetB = WB - FfA = 30.4 N - 60.8 N = -30.4 N
FnetB = mb x ab
Note: Using WB = mb x g: mb = 30.4 N / 9.8 m per sec squared
FnetB = mb x ab
ab = FnetB / mb = -30.4 N (9.8 m/s2) / 30.4 N
ab = -9.8 m/s2
Acceleration direction is upwards. Block B slows down until it stops moving. Distance travelled and duration of motion depends on initial velocity.