Problem: On a graph of acceleration vs force, with acceleration in meters/second/second and force in Newtons, we find the points ( 6, 0) and ( -1, 9). What do the rise and run between these points represent? What does the slope between these points represent?
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Solution: Since the graph is of acceleration vs. force, the dependent variable acceleration will be plotted on the vertical axis with the independent variable force on the horizontal axis. The rise will therefore indicate a change in acceleration while the run will indicate a change in force. The rise represents a change in acceleration from 0 to 9 meters/second/second or an acceleration change of 9 meters/second/second. The run represents a change in force from 6 Newtons to -1 Newtons, which implies a force interval of -7 Newtons. The slope is rise/run = 9 meters/second/second / ( -7 Newtons) = -1.285 (m/s/s)/(Kg m/s/s) = -1.285 /Kg. This 'reciprocal kilogram' might seem not to make much sense. But recall that Newton's Second Law tells us that the acceleration of an object is determined by its net force F, with a = F/m. A graph of a = F/m is a lot like a graph of y = bx, which as you should understand, has slope b. If we write a = F/m in this form as a = (1/m) F, we see that a graph of a vs F could very well have a slope of 1/m. And of course the units of 1/m would be 1/kg.
Generalized Response: On a graph of acceleration v vs. force F, two points will have coordinates (F1, a1) and (F2, a2).
The rise between these points is from a1 to a2, a rise of `da = a2 - a1. This rise represents the difference in acceleration between acceleration a1 and acceleration a2.
The run is from F1 to F2, a run of `dF = F2 - F1. This run represents the difference in force between F1 and F2, or the force difference between F1 and F2.
The slope is the rise divided by the run, which is `da / `dF, the acceleration change divided by the force difference. This is the average rate at which acceleration a changes with respect to force.
For an experiment in which acceleration a is observed as a function of net force F, for example an Atwood machine, we expect from Newton's Second Law that a = F / m. A graph of F vs. a should therefore have slope 1/m, and could be used to infer the mass of the accelerated object.
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Figure description: The graph below shows two points (F1, a1) and (F2, a2) on a graph of acceleration vs. force. The rise is seen to be `da = a2 - a1, representing the change in acceleration. The run is seen to be `dF = F2 - F1, the force difference between the points.
The slope `da / `dF therefore represents the acceleration change divided by the force difference, which is the average rate at which the acceleration changes with respect to force. For a graph of data revealing the acceleration of a given object resulting from various forces, where we expect from Newton's Second Law that a = F / m, the slope of this graph should be 1/m.
