What Do the Slope and Y-Intercept Represent in a Force vs. Acceleration Graph?

AI Thread Summary
In the discussion, participants analyze a Force vs. Acceleration graph derived from an experiment using a cart and pulley system with varying masses. The slope of the graph, calculated from the trend line equation, represents acceleration, as it is derived from the relationship F/m = a, where F is force and m is mass. There is confusion regarding the labeling of the axes, with suggestions that mass should be on the x-axis and force on the y-axis for correct interpretation. The y-intercept of the graph, which is 2.16, could indicate a constant force acting on the system when no mass is applied. Overall, the key takeaway is the importance of correctly identifying variables to accurately interpret the graph's slope and intercept.
yeopar
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Homework Statement



we used a cart and a pulley connected, and we added hangers with different masses. masses on hangers were: 0.02, 0.03,0.04, 0.05, 0.06,0.07 (kg) and the forces measured with GLX machine were: 0.19, 0.29, 0.39, 0.49, 0.59, 0.69. then we found out our acceleration and we graphed everything. Force on y-axis and acceleration on x axis. and we got our equation from the trend line which was: y=2.16+0.11

question: what is the value (including units) and the meaning of the slope of your graph?
what is the value (including units) and the meaning of the Y intercept?






Homework Equations


slope of our graph would mean mass since, F/A=M


The Attempt at a Solution


slope of our graph would mean mass, according to the formula we know derived from Newton's second law of motion, but our masses weren't constant. So what would the slope of the graph mean?

and what does Y intercept of the graph mean?
 
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yeopar said:

Homework Statement



we used a cart and a pulley connected, and we added hangers with different masses. masses on hangers were: 0.02, 0.03,0.04, 0.05, 0.06,0.07 (kg) and the forces measured with GLX machine were: 0.19, 0.29, 0.39, 0.49, 0.59, 0.69. then we found out our acceleration and we graphed everything. Force on y-axis and acceleration on x axis. and we got our equation from the trend line which was: y=2.16+0.11

question: what is the value (including units) and the meaning of the slope of your graph?
what is the value (including units) and the meaning of the Y intercept?






Homework Equations


slope of our graph would mean mass since, F/A=M


The Attempt at a Solution


slope of our graph would mean mass, according to the formula we know derived from Newton's second law of motion, but our masses weren't constant. So what would the slope of the graph mean?

and what does Y intercept of the graph mean?

It looks to me that your data included Force and mass. You have a bunch of masses written down... 0.02kg, 0.03kg, 0.04kg,... these are masses.

So I can only conclude that you were attempting to determine the acceleration which you basically wrote. You graphed Force against mass so the slope you found was acceleration.
F/m = a
 
yeopar said:

Homework Statement



we used a cart and a pulley connected, and we added hangers with different masses. masses on hangers were: 0.02, 0.03,0.04, 0.05, 0.06,0.07 (kg) and the forces measured with GLX machine were: 0.19, 0.29, 0.39, 0.49, 0.59, 0.69. then we found out our acceleration and we graphed everything. Force on y-axis and acceleration on x axis. and we got our equation from the trend line which was: y=2.16+0.11

question: what is the value (including units) and the meaning of the slope of your graph?
what is the value (including units) and the meaning of the Y intercept?






Homework Equations


slope of our graph would mean mass since, F/A=M


The Attempt at a Solution


slope of our graph would mean mass, according to the formula we know derived from Newton's second law of motion, but our masses weren't constant. So what would the slope of the graph mean?

and what does Y intercept of the graph mean?

It looks to me that your data included Force and mass. You have a bunch of masses written down... 0.02kg, 0.03kg, 0.04kg,... these are masses.

So I can only conclude that you were attempting to determine the acceleration which you basically wrote. You graphed Force against mass so the slope you found was acceleration.
F/m = a

I think the part in red I highlighted is where you got a bit off...
 
can i send you the copy of our excel sheet so that you can look it over?
 
*should i?
 
what pgardn is trying to say is that you incorrectly labeled the graph, the independent variable should be the masses you acquired and the dependent variable would be the force.
Now to graph this you'd place mass on the x-axis and force on the y axis. Thus the slope would be f/m=a.
 
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