Consider the system shown in the figure (Figure 1) . Block A weighs 35.8 N and b
ID: 1458903 • Letter: C
Question
Consider the system shown in the figure (Figure 1) . Block A weighs 35.8 N and block B weighs 33.4 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 35.8 N , falls asleep on top of block A. If block B is now set into downward motion, what is its acceleration magnitude?
A cat, also of weight 35.8 N , falls asleep on top of block A. If block B is now set into downward motion, what is its acceleration direction?
Explanation / Answer
Weight of A = W a = Ma * g = 35.8
mass of A = Ma = 35.8/9.81 = 3.649 kgs
Weight of B = Wb = Mb * g = 33.4 N
mass of B = Mb = 33.4/9.81 = 3.4 Kgs
apply Frictional Force Ff = u Wa = T
u = 33.4/35.8 = 0.932
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use Net force T- Ff = ma a + 2Mb a
accleration a = (33.4 - 0.932 * 2* 35.8)/( 3.649 + (2 * 3.4))
a = 3.18 m/s^2