Rand Prob 11

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course Phy 121

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 ( 20 cm/s2) / clip, and if the slope of a graph of number of paper clips needed to maintain equilibrium vs. ramp slope is 47 clips / unit of ramp slope, then how many cm/s2 of acceleration should correspond to 1 unit of ramp slope?  If 52 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?I remember something about Fnet = for small slopes being equal to m * slope(less than .1). It seems that both graphs are graphs of net force( a vs m for the first and m vs slope for the second), or rate of change of a with reference to m for the first and rate of change of mass with reference to slope for the second. In other words, for a change in mass the rate of change in a is equal to the slope of the line, and for every change in slope the mass is equal to the rate of change in the slope of the graph(or change in mass / change in ramp slope). Where in both cases the x axis is independent of the y.

So for the first graph, acceleration changes by 20cm/s^2 for each paperclip added(or when we add the mass of the paperclip to the system). For the second graph is telling me that we need 47 paperclips to maintain 0 acceleration(equilibrium) so for 47 clips @ 20cm/s^2 per clip, each unit of ramp slope would equal 2.35cm/s^2.

Fnet = 47clips * 2.35cm/s^2 = 110.45 Newtons

a = 110.45 Newtons / 52 clips = 2.1cm/s^2

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47 clips, each corresponding to 20 cm/s^2 per clips, would be associate with an acceleration of

47 clips * (20 cm/s^2) / clip = 940 cm/s^2.

To divide 47 clips by (20 cm/s^2) / clip would give you units of s^2 / (cm * clip^2).

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