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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.
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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: ->->->->->->->->->->->-> :
F = 5kg * 9.8m/s^2 = 49 Newtons
F = 6 kg * 9.8m/s^2 = 58.8 Newtons
net force in the positive direction is 58.8 Newtons and that a force of 49 Newtons acts in the negative direction,
so that the net force on the system is 58.8 Newtons - 49 Newtons = 9.8 Newtons.
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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: ->->->->->->->->->->->-> :
fnet = m * a
= 5kg * 1.8 m/s
= 9 Newtons the direction will be negative
1.8 m/s is not an acceleration; a rate of descent is a speed or velocity
acceleration also has units of m/s^2, not 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: ->->->->->->->->->->->-> :
they are in the same direciton because the velocity will be increasing as will the acceleration.
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15 min
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Good start, but you will want to revise the last half.
&#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. &#