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phy 201
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: ->->->->->->->->->->->-> :
5kg x 9.8m/s^2= 49 N - positive
6 kg x -9.8 m/s^2 = -58.8N - negative
-58.8N+49N = -9.8N
9.8N / 11 kg = .89m/s^2
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@& Very good. However you also do need to specify the direction of the acceleration.*@
• 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: ->->->->->->->->->->->-> :
5kg x 1.8 meters/s = 9N
@& You don't get a force when you multiply mass by velocity. 1.8 m/s is a velocity, not an acceleration.*@
The direction will be neative
@& You need to specify at the beginning which direction is positive.
My guess is that you are assuming the positive direction to be the one in which the greater mass descends, in which case you are right that the direction of the initial velocity would be negative.*@
5kg /9N= .55m/s^2
@& RIght idea, but the entire system has a mass of 11 kg, not 5 kg.
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.55m/s^2 x 1 s = .55 m/s
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@& With the correct acceleration, this would give you the change in velocity. However this would not be the velocity after 1 second.*@
• 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: ->->->->->->->->->->->-> :
The first second the velocity and acceleration are the same. The system is in the same direction. The system will speed up
.55m/s^2 x 2sec = 1.10 m/s
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*#&!*#&!*#&!
@& You're getting there, but you should submit a revision so we know you are seeing everything correctly.
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new 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. &#
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