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phy 231
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: ->->->->->->->->->->->-> :
Have a total of 11 kg for system and the side with the 6kg weighs more than the side with 5kg. This means the 5 kg will be negative.
6kg - 5kg = 1 kg
1 kg * 9.8m/s^2 = 9.8 N
OR individually 5 kg * 9.8 m/s^2 = 49 N and 6 kg * 9.8 m/s^s = 58.8 N. Since the forces are acting opposite one another Fnet = 58.8 N - 49 N = 9.8 N
The acceleration resulting is a = F/m = 9.8 N/11 kg = .89 m/s^2
Direction will be towards the 6 kg mass, 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: ->->->->->->->->->->->-> :
We know that a = .89 m/s^2 from the previous problem and v0 = -1.8 m/s. If the pulley comes to a stop and reverses direction because the weight is greater on the other side, the vf = 0. Therefore `dv = (0 - -1.8 m/s) = +1.8 m/s and vAve = (-1.8 m/s + 0)/2 = -.9 m/s
The time it comes to a stop is at `dt = `dv/a = +1.8 m/s/.89 m/s^2 = 2.02 sec
So, reasoning will tell us that if its speed is at 1.8 m/s and we know `a will be 9.8 m/s^2, then:
v(1) = - 1.8 m/s + .89 m/s^2(1 sec) = -.91 m/s after 1 second in the direction of the 5 kg mass, or negative direction.
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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: ->->->->->->->->->->->-> :
The velocity is in the negative direction and the acceleration in the positive direction and the system is slowing down getting ready to reverse directions.
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45 min
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July 17th, 4:02 pm
@& Looks good. Compare your results to the discussion at the link below. No need for any revision unless you have questions. *@
&#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
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