For each of five trials, find the acceleration from ax= (twice distance glider moves)/(time squared) The glider & hanging mass have the same acceleration, and we will call to the left the +x direction for the glider, and down the +x direction for the hanging mass. Using Microsoft Excel, plot a graph of the acceleration vs. the size of the hanging mass. Make sure the graph contains all of the features it should. This graph should have points that follow a straight line. Add a linear trendline to the graph and also display the equation for this line on the graph. We can show that Newton’s Laws predict that the acceleration should be ax=m2g/m1+m2 (3) where m1 is the glider’s mass and m2 is the amount of hanging mass. Equation (3) tells us that the predicted slope of your a x vs. m2 graph should be equal to g/(m1+ m2). Calculate the predicted value of the slope, and then find the percent error between it and the experimental value of the slope. When you do the % error calculation, treat the predicted slope as the “accepted value”. Note: The masses must be in kilograms when you find the predicted value of the slope. We also have The size of the tension force on glider:T1= m1ax (4) and the size of the tension force on hanging mass: T2= m2(g - ax) (5) SinceT1 and T2 are part of an action-reaction pair of forces, the should be the same size. For each of the five trials in Part Two, calculate the values of the tensions and then find the percent difference between the values of T1 and T2. Note: The masses must be in kilograms when you calculate the tensions. The Data is all the information that is provieded.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Related questions

Question

For each of five trials, find the acceleration from ax= (twice distance glider moves)/(time squared) The glider & hanging mass have the same acceleration, and we will call to the left the +x direction for the glider, and down the +x direction for the hanging mass.

Using Microsoft Excel, plot a graph of the acceleration vs. the size of the hanging mass. Make

sure the graph contains all of the features it should.

This graph should have points that follow a straight line. Add a linear trendline to the graph and also display the equation for this line on the graph.

We can show that Newton’s Laws predict that the acceleration should be

ax=m2g/m1+m2 (3)

where m1 is the glider’s mass and m2 is the amount of hanging mass.

Equation (3) tells us that the predicted slope of your a x vs. m2 graph should be equal to g/(m1+ m2).

Calculate the predicted value of the slope, and then find the percent error between it and the experimental value of the slope. When you do the % error calculation, treat the predicted slope as the “accepted value”. Note: The masses must be in kilograms when you find the predicted value of the slope.

We also have

The size of the tension force on glider:T1= m1ax (4)

and the size of the tension force on hanging mass: T2= m2(g - ax) (5)

SinceT1 and T2 are part of an action-reaction pair of forces, the should be the same size.

For each of the five trials in Part Two, calculate the values of the tensions and then find the percent difference between the values of T1 and T2. Note: The masses must be in kilograms

when you calculate the tensions.

The Data is all the information that is provieded.

**Experimental Data Table: Physics Experiment on Mass and Motion**

This table displays the results from a series of five trials conducted to investigate the relationship between mass, distance, and time.

- **Trial**: Numbered from 1 to 5.
- **Total Mass**: Constant at 400 grams for each trial, indicated in red.
- **Hanging Mass**: Incrementally increases by 20 grams per trial, starting at 20 grams and reaching 100 grams.
- **Glider Mass**: Starts at 380 grams and decreases by 20 grams each trial, reaching 300 grams. This is highlighted in red for clarity.
- **Distance**: The distance covered in each trial increases from 40.8 centimeters to 48.7 meters.
- **Time**: The time taken for each trial decreases from 1.45 seconds to 0.69 seconds.

This dataset can be used to analyze how variations in the hanging mass and glider mass affect the distance traveled and the time taken to cover that distance.

Expert Solution

Step 1

Given

$a_{x} = \frac{2 S}{t^{2}} a_{x} = (\frac{m_{2}}{m_{1} + m_{2}}) g T_{1} = m_{1} a_{x} T_{2} = m_{2} (g - a_{x}) T_{1} = T_{2}$

Trial	Total mass m₁ + m₂ (kg)	Hanging Mass m₂ (kg)	Glider Mass m₁ (kg)	Distance S (m)	Time T (s)
1	0.4	0.02	0.38	0.408	1.45
2	0.4	0.04	0.36	0.410	1.18
3	0.4	0.06	0.34	0.42	0.79
4	0.4	0.06	0.32	0.426	0.68
5	0.4	0.10	0.30	0.487	0.69