What should be on the x and y axis of this graph?

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The discussion focuses on plotting a graph of acceleration versus net force, with acceleration on the vertical axis and net force on the horizontal axis. Participants clarify how to calculate net force using the formula Fnet = ma, factoring in applied and friction forces. The experiment involves a cart with weights and a pulley system, emphasizing the importance of maintaining constant mass. The calculations provided demonstrate the relationship between net force and acceleration, indicating a direct proportionality. Overall, the thread seeks to understand the graph's implications on the dynamics of force and acceleration.
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The question is:

Plot a graph of the acceleration (vertical axis) as a function of the net force for the trials in which the mass remained constant. Draw a line of best fit and calculate its slope. What does the graph indicate about the relationship betweena cceleration and net force?

What should be on my x and what should be on my y axis?
 
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acceleration (vertical axis)---y
net force for the trials---x
 
How do I figure out net force?
The experiment is a cart with weights with a string attatched to it, the string dangles over a pulley that is attatched to a flat desk and the weights on the end of the string use the force of gravity to pull the cart forward until the weight on the end of the string touches the floor.http://img210.imageshack.us/img210/6541/untitledab4.jpg
That is what the experiment looks like, and this is what my chart looks like:
http://img338.imageshack.us/img338/7705/adadwr0.jpg
Note, total mass factors in the weights on the string too.So far I have done:

THIS IS FOR THE FIRST ROW
Fa = applied force
Ff = friction force
m = mass (kg)
a = acceleration (m/s^2)Fa + Ff = ma
Ff = ma - Fa
Ff = (.557)*(.43) - .39
Ff = -0.15049

Fnet = Ff + Fa
Fnet = 0.23951Is this correct?
Also how do you figure out Net force to mass? I did this for the first row netforce/kg one.
Fnet = ma
Fnet = .557*.43
Fnet = .23951

So the mass of the first one is .557.
Newtons/kg
.23951/.557
.43 N/kg
 
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