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Phy 231
Your 'cq_1_12.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Masses of 5 kg and 6 kg are suspended from opposite sides of a light frictionless pulley and are released.
• What will be the net force on the 2-mass system and what will be the magnitude and direction of its acceleration?
answer/question/discussion: ->->->->->->->->->->->-> :
For 5 kg side:
F = m*a; m = 5 kg; a = 9.8 m/s^2
F = (5 kg)*(9.8 m/s^2) = 49 N
For 6 kg side:
F = m*a; m = 6 kg; a = 9.8 m/s^2
F = (6 kg)*(9.8 m/s^2) = 58.8 m/s^2
F net = 58.8 N – 49 N = 9.8 N
a = 9.8 N / 11 kg = 0.891 m/s^2
The system will accelerate in the direction of 6 kg mass and its acceleration will have a magnitude of 0.891 m/s^2
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• If you give the system a push so that at the instant of release the 5 kg object is descending at 1.8 meters / second, what will be the speed and direction of motion of the 5 kg mass 1 second later?
answer/question/discussion: ->->->->->->->->->->->-> :
`dt = 1 s; v0 = 0; vf = -1.8 m/s
a = (vf – v0) / `dt
a = (-1.8 m/s – 0m/s) / 1 s = -1.8 m/s^2
vAve = `ds / `dt = (vf + v0) / 2
(2*`ds) / `dt = (v0 + vf)
`ds = [(v0 + vf)*(`dt)] / 2
`ds = [(0 m/s + (-1.8 m/s))*(`dt)] / 2
`ds = -1.8 m/s / 2
`ds = -0.9
Speed = distance / time
Speed = 0.9 m / 1 s = 0.9 m/s because speed cannot be negative. The acceleration is moving in the downward direction on the 5 kg side.
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• During the first second, are the velocity and acceleration of the system in the same direction or in opposite directions, and does the system slow down or speed up?
answer/question/discussion: ->->->->->->->->->->->-> :
During the fist second, the velocity and acceleration of the system are in the same direction and the system is slowing down.
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About 30 minutes
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&#Your work looks very good. Let me know if you have any questions. &#