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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.
** 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: ->->->->->->->->->->->-> :
Fnet = m6 * g + m5 * g = 6 kg * 9.81 m/s^2 + (-5kg 8 9.81 m/s^2) = 9.81 N
The direction will be that of the greater mass of 6 kg.
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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: ->->->->->->->->->->->-> :
0.819 m/s
v0 = 1.8 m/s, a = fnet / mass = 9.81 kg *m/s^2 / 5 kg = -1.962 m/s^2 'dt= 1 s
Fnet = m * a
- 9.81 kg * m/s^2 = 5 kg * a
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The 6 kg mass is also accelerating, so the total 11 kg mass has to be used along with the net force in calculating the acceleration.
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a = -1.952 m/s^2
vf = vo + a 'dt = 1.8 m/s + (-1.952 m/s^2)(1s) = -0.152 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: ->->->->->->->->->->->-> :
During the first second, at v0 and 'dt = 0, velocity and acceleration have different signs, and the system is slowing down.
at 'dt = 0.92 s, the system changes direction and velocity and acceleration are in the same direction.
vf = vo + a*'dt
'dt = vo / a = 0.92 s
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12 mins
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Good, except that you need to use the entire mass of the accelerating system. Check my note.
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