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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.
** CQ_1_12.1_labelMessages **
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30-40 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: ->->->->->->->->->->->-> : Fheavy = 6kg*9.8m/s^2 = 58.8 Newtons. Flight = 49 Newtons. Fnet = 58.8-49 = 9.8 Newtons. 9.8N/11kg = 0.89m/s^2 (accleration in positive
direction or direction of heavy weight/down).
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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: ->->->->->->->->->->->-> : 9.8N/11kg = 0.89m/s^2. Vf = sq root -1.8^2 + 0.89m/s^2*1s = 1.53m/s. ???Did I do this correctly? Always feel I don't get the directions and
negative/positive right???
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You need to start by declaring which direction is positive. This is necessary to avoid confusion, and helps clarify the directions of the various quantities throughout the problem.
By using the positive result .89 m/s^2 as the acceleration, you have implicitly chosen the direction in which the system would naturally accelerate (this being the direction in which the 6 kg mass descends) as positive. Thus the direction in which the 6 kg mass descends must be regarded as positive throughout.
It would have been wise to state clearly at the beginning that the positive direction is the one in which the 6 kg mass descends.
The initial velocity is in the direction in which the 5 kg mass descends, so is negative.
Thus acceleration is .89 m/s^2 and the initial velocity is -1.8 m/s.
This is in agreement with the signs you have used for these quantities.
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You appear to have used an incorrect relationship to find vf.
vf = +- sqrt(v_0^2 + 2 a `ds).
You appear to have used a `dt instead of 2 a `ds. This causes an inconsistency in units.
You also appear to have squared the 1.8 m/s but not the - sign. You have to square the entire expression. This would lead to a final velocity greater than the initial velocity.
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It is important to be able to use the equations of motion correctly, and to use units throughout.
However in this case the reasoning can be much simpler and more insightful. An acceleration of .89 m/s^2 means that the velocity changes by .89 m/s every second.
The result you get from applying this insight will be consistent with the result you get from solving the equations. It would therefore be appropriate to answer the question in both ways. In the process you would deepen your insight into the physics, the equations and the reasoning process.
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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: ->->->->->->->->->->->-> : I think the descending of the light side would make velocity in that direction negative (also because the other heavy side would be
ascending, making v0 -1.8m/s. Acceleration would be positive because it's lighter, it would start to slow down in the direction opposite in which it is pulled (or
opposite the -1.8m/s velocity). So, in summary, and it slows down.
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Good, but again nothing in the statement of the problem or in the situation dictates which direction is positive.
You could equally well have chosen the direction of the initial velocity as positive, making the direction of the acceleration negative.
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The point is that the choice of positive direction is completely arbitrary. It is for this reason that it must be declared at the very beginning of the solution.
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See if you can modify your solution according to my notes.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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