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Phy 121
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 **
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: ->->->->->->->->->->->-> :
The first mass will have a Fnet = 5 kg * 9.8 m/s^2 = 49 Newtons
The second mass will have a Fnet = 6 kg * 9.8 m/s^2 = 58.8 Newtons
The net force of the entire system is Fnet = 49 N + 58.8 N = 107.8 Newtons
a = 107.8 kg * m/s^2 / 11 kg = 9.8 m/s^2
The direction of the acceleration is probably downward.
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If the weights were equal the system would clearly not move. However the same analysis you have applied here would still show that they have an acceleration of 9.8 m/s^2.
Clearly gravity pulls on this system in two opposite directions, though both pulls are downward.
How can we reconcile this?
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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: ->->->->->->->->->->->-> :
We already said the a = 9.8 m/s^2 and since we know this we can calculate the other variables
we know a, v0, `dt, and we are wanting to find vf
vf = v0 + a `dt = 1.8 m/s + 9.8 m/s^2 * 1 s = 1.8 m/s + 9.8 m/s = 11.6 m/s
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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 that acceleration is in the same direction as velocity and that the systme is speeding up.
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20 minutes
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Check my note above and give it a few minutes' thought before looking at the link given below.
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&#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. &#