phy201
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: ->->->->->->->->->->->-> :
Fnet = (5kg + 6kg) ( 9.8m/s^2) = 107.8N
the 6kg object would be accelerating 9.8m/s^2 in the downward direction, reacting to the acceleration of gravity and the 5kg object would be accelerating upwards, reacting by the pull of the 6kg object
Gravity pulls this system in opposite directions.
#$&*
If you give the system a push so that at the instant of release the 5 kg object is descending at 1.8 meters / second, what will be the speed and direction of motion of the 5 kg mass 1 second later?
answer/question/discussion: ->->->->->->->->->->->-> :
vf=v0+a `dt
vf= 1.8m/s +9.8m/s^2+ 1s= 11.6 m/s^2 in the downward direction
#$&*
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 velocity and acceleration of the 5kg side would be in the opposite directions because the velocity would be trying to pull the object down but gravity would be pulling the opposite object down since it is heavier. This would make the system slow down
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15 min
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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.