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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: ->->->->->->->->->->->-> :
Force for 5 kg mass: 5 kg * 9.8 m/s^2 = 49 N
Force for 6 kg mass: 6 kg * 9.8 m/s^2 = 58.8 N
Fnet = 58.8 N - 49 N = 9.8 N
a = 9.8 N / 11 kg
a = .89 m/s^2
The net force is 9.8 N.
The magnitude of the acceleration is .89 m/s^2
The direction will be toward the 6 kg side unless there is a force to change that.
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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: ->->->->->->->->->->->-> :
Force for 5 kg mass: 5 kg * 9.8 m/s^2 = 49 N
Force for 6 kg mass: 6 kg * 9.8 m/s^2 = 58.8 N
Fnet = 49 N - 58.8 N = -9.8 N
a = -9.8 N / 11 kg
a = -.89 m/s^2
`ds = v0`dt + .5a`dt^2
`ds = (1.8 m/s)(1 s) + .5(-.89 m/s^2)(1 s)^2
`ds = 1.8 m + (-.445 m)
`ds = 1.355 m
vf^2 = v0^2 + 2a`ds
vf^2 = (1.8 m/s)^2 + 2(- .89 m/s^2)(1.355 m)
vf^2 = 3.24 m^2/s^2 + (-2.4119 m^2/s^2)
vf^2 = .83 m^2/s^2
vf = .9 m/s
It will be traveling at the rate of .9 m/s downward.
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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: ->->->->->->->->->->->-> :
During the first second, the velocity of the system is positive (or pulling toward the 5
kg weight) but the acceleration is negative (slowing it down because of the force of the
6 kg weight). The system goes from 1.8 m/s to .9 m/s, so it is slowing down. This is
also evidenced by the negative acceleration.
&#This looks very good. Let me know if you have any questions. &#