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course Phy 231
5/3 10I am not sure i have the right method of thinking for this question, and tried to relate the graph of 'dAcceleration per 'dNumberofclips; with the number of clips for one unit of slope.
The term equilibrium, does this refer to the amount of clips needed to have the same value of acceleration of 16 cm/s^2?
Here is my best shot at the question." "If the slope of a graph of the acceleration of a cart vs. the number of paper clips attached by a string and suspended over a pulley is ( 16 cm/s2) / clip, and if the slope of a graph of number of paper clips needed to maintain equilibrium vs. ramp slope is 68 clips / unit of ramp slope, then how many cm/s2 of acceleration should correspond to 1 unit of ramp slope? If 70 clips are necessary to match the mass of the cart, then if we could apply this force to the cart without the extra mass of all those clips, what would be the acceleration of the cart?
Im having trouble with the second part of this question, but i understand that:
there is a increase, or change in acceleration of the system for every one clip, according to the graph, a ratio i record as:
Acceleration / number of clips = 16 cm/s^2 / 1 clip
The second part of this question kind of confused me, but i believe, and correct me here if i am wrong that:
acceleration / number of clips = 1 unit of slope / 68 clips.
So the acceleration from a 1 unit change in slope is = 68 clips * (16 cm/s^2 / 1 clip) = 1088 cm/s^2
The acceleration of the cart at 70 clips hanging and none on the cart = 70 clips * (16cm/s^2 / 1 clip) = 1120 cm/s^2.
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Your reasoning is very good. Nice job.
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