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Phy 231
Your 'cq_1_09.1' 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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About 45 minutes
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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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You didn't show the details of your calculations, but everything appears to be OK. Check against the link and let me know if you have questions; otherwise nothing more is required.
&#See any notes I might have inserted into your document. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#