Consider the system shown in the figure (Figure 1) . Block A weighs 42.2N and bl
ID: 1280386 • Letter: C
Question
Consider the system shown in the figure (Figure 1) . Block A weighs 42.2N and block B weighs 32.5N . Once block B is set into downward motion, it descends at a constant speed.
A cat, also of weight 42.2N , falls asleep on top of block A. If block B is now set into downward motion, what is its acceleration magnitude??
Consider the system shown in the figure (Figure 1) . Block A weighs 42.2N and block B weighs 32.5N . Once block B is set into downward motion, it descends at a constant speed. A cat, also of weight 42.2N , falls asleep on top of block A. If block B is now set into downward motion, what is its acceleration magnitude??Explanation / Answer
"Once block B is set into downward motion, it descends at a constant speed."
Therefore, the tension on the cord due to the weight if block B is equal to the kinetic friction force between block A and the tabletop.
In fact, the kinetic friction force is 17.724 Newtons.
42.2 * 0.420 = 17.724
When the cat adds it's weight to block A, the kinetic friction force instantly doubles.
The net force on the system is 17.724 Newtons.
17.724 * 2 - 32.5 = 2.948 N
The weight of the system is 116.9 Newtons, or 11.93 Kilograms.
42.2 * 2 + 32.5 = 116.9
116.9/ 9.8 = 11.93
So, the acceleration on the system is 1.699 m/s/s.
17.724 / 11.93 = 1.49
The system quickly comes to a stop.