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:
Answer: F = m*a = 1kg * 9.8m/s^2 = 9.8N in the direction of acceleration of gravity on the 6kg mass side. The acceleration will be 9.8m/s^2
• 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:
Answer: v0 = 1.8m/s, a= -9.8m/s^2, dt = 1s, vf = v0 + adt = 1.8m/s + (-9.8m/s^2)(1s) = -8m/s. The speed will be 8m/s going in the direction of gravity which will be pushing the 6kg side downward and the 5kg side upward.
• 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:
Answer: The velocity starts at 1.8m/s on the lighter side, then increases on the heavier side. The acceleration of the object is pushing against the direction of motion upon first release then accelerates with the object as momentum switches over to the heavier side. The system speeds up.
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20 minutes
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Let the positive direction be the one in which the 6 kg mass is descending. Note that the weights of the masses are about 59 N and 49 N.
Two forces thus act on the system, the 59 N force in the positive direction and the 49 N force in the opposite direction, giving us a net force of about
F_net = 59 N - 49 N = 10 N.
The mass of the system is 11 kg. A net force of 10 N on a system of mass 11 kg results in acceleration
a = F_net / m = 10 N / (11 kg) = .9 m/s^2, approximately.
If the 5 kg mass is descending at 1.8 m/s, then the velocity of the system is -1.8 m/s (note that all directional quantities must be referenced to our original choice of positive direction).
An acceleration of .9 m/s^2 means that in 1 second the velocity of the system will change by `dv = a `dt = .9 m/s^2 * 1 s = .9 m/s. This will give us a velocity of vf = v0 + `dv = -1.8 m/s + .9 m/s = -.9 m/s. That is, the 5 kg object will still be descending, but at .9 m/s rather than at 1.8 m/s.
During this 1-second interval the acceleration is positive and the velocity remains negative. So velocity and acceleration are in opposite directions. Whenever this is the case the object is slowing down, as it clearly is in this example.
If another second passes then the object's velocity will be near zero. After reaching the state of rest for an instant, the continuing acceleration will result in a positive velocity (6 kg mass descending), which with the positive acceleration will then begin speeding the system up.