ttached to 1.63 ko that is initially at rest on a frictionless table. A third ma
ID: 1644944 • Letter: T
Question
ttached to 1.63 ko that is initially at rest on a frictionless table. A third mass-2.41 kg. which is alo initially at rest on a trictionless table, is attached to the to A hanging mass, - 2.39 kg, is attached by a gnt string that nuns over a frictionless pulley to the front af a mass M hackar M2 by light string. atachsd ) Find the msunitude of the acceleration, a, of mass s Subinit Answer Tries us ind the and Mz. b) Find the tension in the string between masses Submit Answer Tries /5 Submit Aner Tries 05Explanation / Answer
Mass accelerating the system is 2.38 kg. This will exert a force of
F = ma newtons, where m = 2.38, a = acceleration due to gravity = 9.81 m/s^2
So
F = 2.38 * 9.81 = 23.35 N
The total mass that is being accelerated is
2.38 + 1.63 + 2.41 = 6.42 kg
A)
The acceleration of the whole system (including M3)
23.347 = 6.42 * a
a = 23.347 / 6.42 = 3.637 m/s^2
B)
The tension in the string between M2 & M3.
F = ma = 3.637 * 2.41 = 8.764 N
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