# Determining what the y-intercept means in a force lab

In summary, the conversation involved a lab experiment where a glider was placed on an airtrack and moved using a pulley system. The mass on the pulley was changed with each iteration and a graph was created to find the mass of the glider. However, the y-intercept was not understood and was predicted to be 1 based on the given equation. The actual y-intercept turned out to be -2.57, leading to a large percent error. The expert provides an explanation that the y-intercept represents the value of g/a for an extremely large mass on the pulley, and suggests thinking about what this value would be intuitively. The expert also points out an error in the graph where the x-values were being

## Homework Statement

In lab, we put a glider on an airtrack and used a pulley to move the glider. The mass on the pulley was changed with each iteration. I have created a graph to find the mass of the glider, but I don't know what the y-intercept is supposed to mean. The graph I have is linear, and the slope is in units of kg.

## Homework Equations

(g/a) = (m1)(1/m2) + 1
(g/a) --> 9.8/acceleration that was found experimentally
m1 --> mass of the glider and the slope of the line
m2 --> mass that was attached to the pulley

## The Attempt at a Solution

Part of the lab is to predict the y-intercept. I predicted that the y-intercept would be 1 based on the equation that we were supposed to use, but the y-intercept ended up being -2.57. I know that leads to a large percent error. What I need to know is what the y-intercept is supposed to represent in this context.

Thanks for any help that you can give!

So, you are plotting g/a on the vertical axis and 1/m2 on the horizontal axis. The y-intercept corresponds to making 1/m2 equal to zero. This would mean making m2 infinite. In other words, the y-intercept is essentially the value of g/a for an extremely large value of m2. Intuitively, what would you expect g/a to be in this case? Thinking about this might help you understand the predicted value of 1 for the y-intercept.

It could be that your experimental value for the y intercept is sensitive to error in your data.

Can you upload a figure showing your graph? If not, can you list your experimental data points (1/m2, g/a)?

I've added a picture of my data as well as the graph.

#### Attachments

• Physics.PNG
20 KB · Views: 833
OK. As you can see from your graph, the difference between +1 and -2.57 for the y-intercept is actually very small considering that your y-data extends over the range of about 8 - 54. So, I think a y-intercept of -2.57 can be considered "close" to a y-intercept of +1 even though the % difference of these two numbers is large.

I noticed that you apparently used the values for ##\frac{1}{m_2}+1## rather than the values of ##\frac{1}{m_2}## for your "x" values for the graph. Did you mean to do that?

The equation in the lab manual is g/a = m_1*(1/m_2) +1. Given this, figured that I should be adding 1 to my x values, but now I'm thinking that shouldn't be doing that.
Also, I apologize for the awful formatting. I'm still trying to figure out how to format equations.
I just changed my graph so that the data used for the x-values didn't include the +1, and my y-intercept is now -1.2

The equation in the lab manual is g/a = m_1*(1/m_2) +1. Given this, figured that I should be adding 1 to my x values, but now I'm thinking that shouldn't be doing that.
Also, I apologize for the awful formatting. I'm still trying to figure out how to format equations.
I just changed my graph so that the data used for the x-values didn't include the +1, and my y-intercept is now -1.2
The equation for a straight line on x-y axes with y-intercept b is y = mx + b. A graph of this line would plot y on the vertical axis and x on the horizontal axis. It would not plot x + b on the horizontal axis.

Likewise, if you have (g/a) = m1(1/m2) + 1, this would give a straight line with y-intercept +1 if you plot (g/a) on the vertical axis and 1/m2 on the horizontal axis. You do not want to plot 1/m2 + 1 as the "x" values.

So, making this change does help a little in moving the y-intercept up. I think a value of -1.2 is actually very good compared to the expected value of +1.

Thank you so much for all of your help! I really appreciate you taking the time to explain it to me.

You are welcome. Enjoy your course.

## 1. What is the importance of determining the y-intercept in a force lab?

Determining the y-intercept in a force lab is important because it gives us information about the initial force or starting point of a system. It helps us understand the relationship between the independent variable (usually force) and the dependent variable (usually displacement or acceleration).

## 2. How is the y-intercept calculated in a force lab?

The y-intercept is calculated by finding the point where the regression line intersects the y-axis. This is usually done using a graphing software or by hand using the slope-intercept formula (y = mx + b) where b represents the y-intercept.

## 3. Can the y-intercept be negative in a force lab?

Yes, the y-intercept can be negative in a force lab. This indicates that the system had an initial force in the opposite direction of the force being applied. It is important to consider the sign of the y-intercept when interpreting the results of a force lab.

## 4. What does a larger y-intercept value indicate in a force lab?

A larger y-intercept value indicates a greater initial force or starting point of the system. This could be due to factors such as the mass of the object or the strength of the applied force. It is important to analyze the y-intercept in relation to the other data points on the graph to understand the overall behavior of the system.

## 5. How does the y-intercept relate to the slope of the regression line in a force lab?

The y-intercept and the slope of the regression line are closely related in a force lab. The y-intercept represents the initial force, while the slope represents the rate at which the force changes with respect to the displacement or acceleration. Together, they give us a complete picture of the relationship between force and the dependent variable.

• Introductory Physics Homework Help
Replies
5
Views
3K
• Introductory Physics Homework Help
Replies
15
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
980
• Introductory Physics Homework Help
Replies
30
Views
2K
• Introductory Physics Homework Help
Replies
19
Views
611
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
25
Views
2K
• Introductory Physics Homework Help
Replies
33
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
962