Block B hangs from Block A by a Rope 1. Rope 2 hangs below Block B.
Each block has a mass of 1.0 kg.
Each rope has a mass of 250 g.
The entire assembly is accelerating upward at 3.00 m/s^2 by force F.
A. What is F?
B. What is the tension at the top of Rope 1?
C. What is the tension at the bottom of Rope 1?
D. What is the tension at the top of Rope 2?
F = ma
T = Sum(forces acting on that point)
The tension force at any point in the system = sum of forces acting at that point.
The Attempt at a Solution
The total mass of the system is 2 x 1000 g + 2 x 250 g = 2500 g
The net acceleration on the entire system = 3.00 m/s^2. So the force at any point is that acceleration x the mass below that point.
A. F = ma = 2500 g x 3.00 m/s^2 = 7500 g-m/s^2 = 7.4 N
B. F = ma = (250 g + 1000 g + 250 g) x 3.00 m/s^2 = 4500 g-m/s^2 = 4.5 N
C. F = ma = (1000 g + 250 g) x 3.00 m/s^2 = 3750 g-m/s^2 = 3.75 N
D. F = ma = 250 g x 3.00 m/s^2 = 750 g-m/s^2 = 0.75 N
And the tension on the bottom of Rope 2 = 0 N.
Did I get that right?
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