Consider the system shown in the figure (Figure 1) . Block A weighs 37.5 N and b
ID: 1531495 • Letter: C
Question
Consider the system shown in the figure (Figure 1) . Block A weighs 37.5 N and block B weighs 18.9 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 37.5 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 37.5 N , falls asleep on top of block A. If block B is now set into downward motion, what is its acceleration direction? (Upwards or Downwards)
A BExplanation / Answer
Here ,
WA = 37.5 N
WB = 18.9 N
a) let the coefficient of kinetic friction between A and the table top is u
as the block is moving with constant speed
tension in the string is T
T - wB = 0
T = 18.9 N
balancing the horizontal force on A
T - u * WA = 0
18.9 - u * 37.5 = 0
solving for u
u = 0.504
the coefficient of kinetic friction is 0.504
b)let the acceleration of the block is a
a = net force/total mass
a = (0.504 * (37.5 + 37.5)) * 9.8/(37.5 + 37.5 + 18.9)
a = 3.95 m/s^2
the acceleration magnitude is 3.95 m/s^2
part C)
the direction of acceleration is upwards