# Force & Acceleration graph

1. Dec 31, 2009

### yeopar

1. The problem statement, all variables and given/known data

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?

2. Relevant equations
slope of our graph would mean mass since, F/A=M

3. 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?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 31, 2009

### pgardn

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

3. Dec 31, 2009

### pgardn

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...

4. Dec 31, 2009

### yeopar

can i send you the copy of our excel sheet so that you can look it over?

5. Dec 31, 2009

### yeopar

*should i?

6. Dec 31, 2009

### jkerrigan

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.