Evan,+Sam,+Erica

= = Lab: Acceleration Down an Incline = Members: Sam Fihma, Erica Levine, Evan Bloom = = Period 2 = = Due Date: December 14, 2010 = The purpose of this experiment was to determine the coefficient of kinetic friction between the metal ramp and the wooden block.
 * Purpose**

If we can determine the acceleration of the block down the ramp, then we can solve for the coefficient of friction between the block and the ramp.
 * Hypothesis**

Picture of setup of the experiment media type="file" key="movierampyofriction.mov" width="300" height="300" Video of a trial of the experiment
 * Procedure**

We measured the degrees of the ramp and calculated the acceleration of the mass three times. We found the average acceleration for each angle and then doubled it due to reasons explained in the conclusion. The sine of the angle was recorded also in order to make the graph below.



You can find the graph and table here: = =

Percent Error of Our Slope compared to Gravity:





This sheet shows our free body diagrams, the theoretical acceleration, and what our graph means. It also includes the calculations to find our coefficient of friction and the percent error between our coefficient and last week's number.

The average slope for both classes was 10.443. The Percent Difference between the class average and our result was calculated.
 * Name || Slope || y-intercept || R2 || slope with y-int = 0 ||
 * Rachel || 10.398 || -1.9849 || 1 || 3.99 ||
 * Evan || 10.476 || -1.8539 || 0.9976 || 4.03 ||
 * Nicole || 9.0393 || -1.9593 || 0.978 || 3.72 ||
 * Justin || 8.7236 || -1.7436 || 0.9782 || 2.31 ||
 * Andrew || 13.406 || -3.6733 || 0.9839 || 3.38 ||
 * Ryan || 11.223 || -2.5289 || 0.9696 || 2.63 ||
 * Emily || 10.089 || -2.0541 || 0.9987 || 2.23 ||
 * Jessica || 9.7156 || -1.8744 || 0.982 || 3.68 ||
 * Jae || 10.393 || -2.6813 || 0.9891 || 4.36 ||
 * Deanna || 9.8699 || -2.264 || 0.9888 || 3.01 ||
 * Chris H || 10.089 || -2.0515 || 0.989 || 4.3 ||
 * Eric || 10.0376 || -2.3396 || 0.9951 || 4.21 ||
 * Sam || 12.299 || -3.0632 || 0.9935 || 4.15 ||



Discussion Questions

1. Discuss your graph. What does the slope mean? What is the meaning of the y-intercept? This is a graph of Acceleration vs. Sin of Theta. Our slope is 10.476, which should be equal to the force of gravity, which normally is 9.81. After doing some calculations which can be seen in number 2 in the sheet scanned above, you can see why g (gravity) should be equal to the slope. The Y-Intercept is equal to negative gravity times mew times the cosine of theta. This is also proved in the sheet above. By changing the Y-Intercept, you are directly affecting the slope. Thisis due to the fact that the angle measurement is considered in both the Y-Intercept and slope. When the angle is steeper, the Y-Intercept is smaller and vice versa. SO when you set the Y-Intercept to 0, the means you are dealing with a frictionless surface. And friction would only increase the slope.

2. If the mass of the cart were doubled, how would the results be affected? If the mass of the cart were to be doubled, acceleration, too, would be doubled. F = ma 2mg*sin(theta) - µ2mg*cos(theta) = 2ma 19.6sin(theta) - µ19.6cos(theta) = 2a

3. Consider the difference between your measured value of g and the true value of 9.80m/s^2. Could friction be the cause of the observed difference? Why or why not? The slope of our graph was 10.476, a little bit greater than the exact value of acceleration due to gravity. Friction would slow down the acceleration of the block down an incline, not increase it. Therefore, it is logical to assume that friction was not the cause of the observed difference. In addition, the y intercept value on our graph represents friction divided by mass. Since the value of friction is represented in the y intercept, it clearly does not have any influence over the acceleration of the block, or the slope of the graph.

4. How were your results in Part B? Why was the expectation that your results be within 2% considered be reasonable when in other labs we allow much larger margins of error?

Conclusion

The purpose of our lab was to investigate how varying incline angles affect the acceleration of a wooden block down an aluminum track. We hypothesized that there was a direct relationship between the two based off of the equation, a= (gravity*mass*sintheda- coefficient*mass*gravity*costheda)/mass.F. The data that we collected and the graph that we made from this data support our hypothesis. The slope of our graph was 10.476, indicating that our calculated value of acceleration due to gravity was 10.476 m/s(squared). This slope only had a 6.7% error from the accepted value of acceleration due to gravity (9.8 m/s(squared)), proving our results to be fairly accurate. Yet, there are multiple possible sources of our minimal error. Although the track was cleaned, it is possible that there was dust, dirt, or other foreign objects that may have obstructed the calculated value of acceleration. Additionally, it is possible that the block was not horizontally level. At times we noticed that the block would slide not only down the track, but horizontally across it. This could have contributed to some error in our calculation for acceleration.

We also experienced some error in our procedure that we were able to fix. At first, we collected data for acceleration based off of angles lower than 10. When the slope of our graph turned out to be less than 1, we knew we had made a mistake. All the data that we used with angles lower than 10 were faulty because in our previous lab we found that 10 was the minimum angle for the block to slide at constant speed. After conducting more trials with angles all about 10 degrees, we disregarded the data from all angles under 10 degrees and formed a new graph. This made our data much more accurate, with a slope of 5.24 m/s^2. Yet, this still was not correct because the value of acceleration due to gravity is 9.8 m/s^2. After realizing another procedural error, we doubled our slope to get an acceleration of 10.476 m/s^2, a value fairly accurate to the accepted value. The reason for this doubling is because we calibrated the photogate timers wrongly. We only measured half of a cycle, instead of a full one.

Part B Calculations:

= = = Lab: Coefficient of Friction = = Members: Sam Fihma, Erica Levine, Evan Bloom = = Period 2 = = Due Date: December 14, 2010 =


 * Purpose:** The purpose of this experiment is to measure the coefficients of both static and kinetic friction between surfaces. Also, we are trying to determine the relationship between the friction forces and the normal force.


 * Hypothesis:** If we can determine the normal force of the table on the block and the force of friction along the surfaces of the block and the aluminum track, then we can find the coefficient of friction. If we were to pull a wooden block along the aluminum surface and make a graph of force vs. time, the peak would be tension, and the slope of the graph would be the coefficient of friction, mew.

This chart was assembled when we dragged the block with mass across the track. We did three trials of each weight of the block and found the mean tension when the block was moving at constant speed. We did this by observing the Time v. Force graph in data studio. We averaged all three trials to come up with an average tension for each weight. We knew that this average tension was equal to the force of friction because: On the x-axis, F=ma T-f = ma We know that acceleration is 0 because the block is at constant speed. T=f This is our kinetic force of friction because the definition of it is the force that must be applied to an object to keep it moving at constant speed.
 * Data Tables:**

This chart also uses the same block mass and weight as the chart above. Each of these readings was the highest point on the Time v. Force graph. This is the peak just before the block begins to move. We know that it is equal to force because of the same logic used above. This point represents static force of friction because it is defined as the force that must be overcome in order to start moving an object.


 * Graph:**

You can find our graphs and tables here:

This was our percent difference for our static friction compared to the class average.
 * Percent Difference:**

Percent Difference = (| X1 – X2|) / (| (X1 + X2) / 2 |) x 100

Percent Error = (|0.2137 – 0.2170|) / (| (.2137 + .2170) / 2 |) x 100

Percent Error = 1.39%

This was our percent difference for our kinetic friction compared to the class average.

Percent Difference = (| X1 – X2|) / (| (X1. + X2) / 2 |) x 100

Percent Error = (|0.1849 – 0.1876 |) / (| (.1849 + .1876) / 2 |) x 100

Percent Error = 1.45%


 * Incline Data Charts**

Static Friction In this part of the lab, we put a block on the end of an aluminum track and raised one side of the track until the block started to move. We recorded the angle of the track to the table at the point the block started moving. This represents static force of friction because it is defined as the force that must be overcome in order to start moving an object.

These calculations were done in order to find the coefficient of static friction.

X Axis Fx=max f –wx = ma Acceleration is at 0 because we are working with constant speed f = wx wx = (m)(g)(sin q ) f = (m)(g)(sin q ) f = (.363)(9.8)(sin11.88) f = .732 N

Y Axis Fy=may N - wy = ma Acceleration is at 0 because we are working with constant speed N = wy wy = (m)(g)(cos 0 ) N = (m)(g)(cos 0 ) N = (.363)(9.8)(cos11.88) N = 3.481 N

f = ( m )(N) .732 = m (3.481) m = .210 The static friction coefficient is .210.

Note: 0 = theta and u = mew due to a wiki malfunction.

For this chart, the block was the same in the chart above. What we did here was raised one end of the track to a point where the block still wouldn't move. We then pushed the block lightly and it continued at constant speed. The angles at which the ramp was at were recorded. This is our kinetic force of friction because the definition of it is the force that must be applied to an object to keep it moving at constant speed.

These calculations were done in order to find the kinetic friction coefficient.

X Axis Fx=max f –wx = ma Acceleration is at 0 because we are working with constant speed f = wx wx = (m)(g)(sin 0 ) f = (m)(g)(sin 0 ) f = (.363)(9.8)(sin10.06) f = .621 N

Y Axis Fy=may N - wy = ma Acceleration is at 0 because we are working with constant speed N = wy wy = (m)(g)(cos 0 ) N = (m)(g)(cos 0 ) N = (.363)(9.8)(cos10.06) N = 3.503 N

f = ( u )(N) .621 = u (3.503) u = .177

The kinetic friction coefficient is .177.

When the track was perpendicular to the ground: Static Friction Coefficient: .210 Kinetic Friction Coefficient: .177
 * Summary:**

When the track was on an incline: Static Friction Coefficient: .214 Kinetic Friction Coefficient: .185


 * Percent Differences Between our two results:**

__ KINETIC FRICTION __

Percent Difference = 1.75%
y=mx In our graph, the friction force takes the place of the y value and the normal force takes the place of the x value. The slope is always multiplied by x, and in this case, the coefficient of friction is being multiplied by the normal force (the value of x).
 * Discussion Questions:**
 * 1. Why does the slope of the line equal the coefficient of friction? Show this derivation.**

The coefficient of static friction between wood and aluminum is 0.22 according to Hypertextbook.com (http://hypertextbook.com/facts/2005/wood.shtml). Our experimental coefficient was .2137 which was very close to the one on that website. This shows how close we actually got to the real number. Our percent error is below.
 * 2. Look up the coefficient of friction between wood and aluminum. Discuss if your measured results fall within the range of theoretical values. Be sure to cite your source!**

Percent Error = (|Experimental Value – Theoretical Value| x 100) / Theoretical Value

Percent Error = (|0.2137 – 0.22| x 100)/.22

Percent Error = 2.86%

Acceleration is zero in this lab, which causes the friction force to be equal to the tension force. Therefore, the magnitude of the tension force on the rope directly affects the magnitude of the friction force calculated. In addition the surface the wooden block was being rubbed against affected the magnitude of the friction force. If the surface was wet, dusty, etc., the magnitude of the friction force would have been altered. Both normal force and friction force affected the coefficient of friction in this lab. This influence is derived from the equation,. The equation clearly shows how these variables would affect the coefficient of friction. Acceleration is also zero on the y axis, so the normal force equals the weight of the object. As the weight of the object moving across the surface changes, the coefficient of friction will change in accordance. Additionally, the force of friction, or tension previously mentioned also affects the magnitude of the coefficient of friction.
 * 3. What variables affected the magnitude of the force of friction? What variables affected the magnitude of the coefficient of friction?**

The coefficient of kinetic friction is smaller than the coefficient of static friction. The coefficient of kinetic friction is .1849, whereas the coefficient of static friction is .2137. This makes sense because static friction is the maximum force to get the object to move, so it will always be larger than the coefficient of kinetic friction.
 * 4. How does the value of coefficient of kinetic friction compare to the value for the same material's coefficient of static friction?**

While the change was not drastic, the coefficient of friction did change when the track was on an incline. When the track was flat, the static friction coefficient was .210 and the kinetic coefficient was .177. When the track was on an incline, the static coefficient was .214 and the kinetic coefficient was .185. Since weight was split into two components during the incline, there the normal force of the track on the block decreased, causing a greater coefficient.
 * 5. Did putting the track on an incline significantly change the coefficient of friction? Why or why not?**


 * Conclusion:**

In performing this lab, we were trying to figure out the coefficient of friction between the aluminum track and the wooden block. In addition, we intended to investigate the relationship between the normal and friction forces. Based off of the equation f= m N, we hypothesized that the relationship between normal force and friction force was direct. Our graph for both kinetic and static friction vs. normal force were linear, proving that the relationship between these two forces was indeed direct. As the normal force increased, so did the friction, and vice-versa. We also hypothesized that the coefficient of static friction would be larger than the coefficient of kinetic friction. This hypothesis was validated by the graphs of maximum friction vs. weight and average friction vs. weight that we made based off of our collected data. We found the slope of both of these graphs to be the coefficient of friction. The graph where we used maximum friction gave us the coefficient of static friction, .2137 N. This was larger than the coefficient of kinetic friction, .1849 N, which we got from the slope of the graph with average friction. This data proves our hypothesis correct. Our percent difference from the class average for static friction was 1.39%, and 1.45% for kinetic friction. We also know that our results were fairly accurate because our results from Part A and Part B/C were very similar. In Part B and C, we found another way to calculate the coefficient of static and kinetic friction. We calculated static in Part C, when we saw the angle that made the block first move. We calculated kinetic in part B, when we saw the angle that made the block move at constant speed. The percent difference between our two values for the coefficient of kinetic friction was 4.37%. The percent difference between our two values for the coefficient of static friction was 1.75%. These close results prove the accuracy of our data. Although our collected data was fairly accurate there was room for mistakes in our procedure. It is possible that when we pulled the cart across the aluminum track, it wasn’t exactly at constant speed. It was fairly impossible to get the acceleration to zero without the help of any technology. This source of error could have affected our results significantly. Additionally, any dust, water or other foreign objects could have altered the coefficient of friction It is also possible that the string attached to the wooden block wasn’t pulled at exactly a zero degree angle. This fault would cause changes in the forces in the system. In part B, there was also room for error. It is quite possible that we misjudged precisely when the block was moving at constant speed, or at exactly what angle it started to move. For future experiments it is important to make multiple changes in order to yield more accurate and precise results. It is important to use a device that can ensure we are pulling the block at constant speed. It is also necessary to use a protractor to make sure that we aren’t dragging the black at an angle.

= Lab: Atwood's Machine = = Members: Sam Fihma, Erica Levine, Evan Bloom = = Period 2 = = Due Date: December 7, 2010 =

=
**Purpose:** The purpose of this lab is to prove Newton's Second Law of Motion by showing that force, mass, and acceleration are all related. Also, it is to show the direct relationship between force and acceleration.======

**Hypothesis with Rationale:**

If there is a direct relationship between net force and acceleration, and we test this theory using an Atwood Machine and constant mass, the resulting graph comparing the two variables should be linear, with slope representing mass.

Newton’s Second Law of Motion states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object. The equation generated from this law is: F=ma. This clearly shows that with a constant mass, force and acceleration will increase simultaneously. To test the validity of this law, we kept mass constant and tested for force and acceleration. In doing so, we predict that if the law is correct, a graph of our data will produce a linear graph, indicating that acceleration and force are directly related.

**Procedure:**
 * 1) Set up a string on the wheel of the Atwood's Machine with two equal masses on each end of the string. The machine needs to be clamped onto an elevated pole. You may want to put tape around the weights to keep them from falling off. The machine should be plugged into a data studio on the computer.
 * 2) Add additional mass to one end of the string. Record the total of each of the masses in kilograms
 * 3) Pull the lighter mass down so that it is lower than the larger one.
 * 4) Let go of the lighter mass and find the acceleration using Data Studio (slope of a V-T graph).
 * 5) Repeat Steps 3-4 with the same masses to verify results and ultimately, find an average of your results.
 * 6) Remove some of the additional mass from the one end and add it to the other. Make sure that the total mass doesn't change. Repeat Steps 2-5 with different amounts of mass on each side.
 * 7) Create an acceleration v. force graph and check to see if the slope is equal to the total mass of the two body system. This will be shown below.

This is a picture of Atwood's Machine clamped to the pole and the strings hanging from the wheel. This is a picture of our experiment with Atwood's Machine on top of the pole and the weights hanging by string near the surface of the table.
 * Setup:**

**Data Table:** This is an organized chart of the data we found. In columns A and B, there is the mass of the two hanging masses during each trial. The next column is total mass of the two body system, which stays constant throughout the experiment. The next three columns are observed accelerations when we used the corresponding masses to the left. The acceleration of the masses were found by using the slope of the v-t graph on data studio. The average column is the three trials averaged together to form an assigned acceleration. The experimental force was the force we found using our own numbers and the theoretical acceleration was the acceleration that we should have gotten given ideal conditions (meaning no error or friction).

__Same Calculations:__ Symbols used - MA = Mass of A, MB = Mass of B

Experimental Force Force = (MA x g) - (MB x g) Force = (0.5kg x 9.8m/s^2) - (0.45kg x 9.8m/s^2) Force = 0.49 N

Theoretical Acceleration Acceleration = ((MA - MB) x g) / (MA + MB) Acceleration = ((0.5kg - 0.45kg) x 9.8m/s^2) / (0.5kg + 0.45kg) Acceleration = 0.516 m/s^2

**Graph:** This graph was created taking the average acceleration (x) from our experiment and our calculated experimental force (y). From the trendline, you can see that they have a linear relationship. Both the data table and graph can be found here:

**Analysis Question (1):**

This force v. acceleration graph is linear, showing how force and acceleration have a direct relationship. The graph has a great trendline with an equation of y=1.0196x+0.0117 and a R squared value of 0.9974. The slope of this trendline is 1.0196. The slope corresponds to the total mass in our experiment. The two are not exactly the same, but are pretty close. We ended up having a percent error of 7.3% which is proved below. Friction was the biggest source of error and that is discussed below. The slope is equal to the mass because of the equation F=ma. In the graph, force is the y value and acceleration is the x value. Slope is the number that is multiplied by x and in this case, mass is being multiplied by x. With mass as our constant, this shows how force and acceleration are directly proportional. As force is increased, acceleration does too. As force is decreased, acceleration does too. The y-intercept value of this graph is very close to 0, meaning when you plug in 0 for x (no acceleration), y=0 (no force). This means that when there is no acceleration, there is no force and vice versa. There is no force and acceleration due to the fact that the masses are equal in mass, thus causing no movement.
 * 1. What is the slope of the trendline? To what actual observed value does the slope correspond? How does it compare to this actual observed value (percent error)? Show why the slope should be equal to this quantity. What is the meaning of the y-intercept value?**

**Percent Error:**

**Analysis Questions (2-4):**

**2. What would friction do to your acceleration? Would you need a bigger or smaller force to create the same acceleration? Was your slope too big or too small? Can friction be a source of error in this experiment? Redo the calculation of acceleration WITH friction to show its effect.**

Friction would slow down our acceleration because it is a force that opposes motion. If friction were present, a larger force would be needed to create the same acceleration. This is because friction is subtracted from the force. The equation used is A= m2(g)/ m1 + m2. Subtracting friction from the top would produce a smaller fraction, making it necessary for force to increase to produce the same acceleration. Our slope was too big in comparison to the actual mass we used. This large mass was probably due to the fact that we didn’t take friction into account. If friction was subtracted from the numerator, ia smaller acceleration value would be produced. Our b value in our equation represents friction. Below is a new calculation taking friction into account. A= F/M A= m(g) – f/ m1 + m2 A= .450(9.8) - .0117/ .950 A= 4.3983/9.50 A= .463 m/s^2 The acceleration that we calculated using Data Studio and ignoring friction was .497 m/s^2. This new value taking friction into account is decreased by .034 m/s^2. This lowered acceleration would cause mass to be lowered to a number closer to the mass we actually used while experimenting. Therefore, the slope of our graph would be more accurate. Clearly, friction makes a difference in our calculations.

**3. Discuss the precision of your data.** Our results happened to be very precise. The largest gap that we had between trials was .017 m/s2, which is not very big. The precision is reflected in our R^2 value, which is .9974, and our percent error, which is a little over 5%. The precise results led to accuracy, which, in turn, led to our strong results.

**4. The real pulley and mass arrangement is not as simple as we assumed. In fact the pulley is not massless and frictionless means that it does require a net "torque" (a turning force) to make it rotate-- this is supplied by the tension in the string. The rotational inertia of the pulley then adds an equivalent mass to the total mass being accelerated, where the equivalent mass for the pulley is approximately equal to 1/2 the mass of the pulley. if the mass of each pulley is 5.6 g could the pulley mass account for a significant portion of your error in the experiment?** If we were to take the mass of the pulley into account, it would account for some error in the experiment. However, it would not at all be a significant portion of error. Our total mass would only change by 0.0028 kg, which is a very small number in relation to what we are working with. While our answer would show some difference, it is a very small portion of error.

**Conclusion:**

The purpose of this lab was to prove that that the equation of Newton’s Second Law, F=ma, is valid. We hypothesized that if this equation was valid, a graph of acceleration vs. force would be linear, indicating that the two variables were directly related. We used an Atwood’s Machine to test this hypothesis. We transferred masses from mass A to mass B, in order to keep the mass constant so we could view the relationship between acceleration and force. Using the mass value we used in the experiment and the acceleration we found by taking the slope of the velocity vs. time graph on Data Studio, we used Newton’s Second law equation to calculate force. If this law was truly accurate the force calculated should be in direct relation with the acceleration. Our graph did indeed produce a linear slope. Based off this, we can come to the conclusion that Newton’s Second Law equation is accurate. The slope of our graph represented the total mass we used in the experiment. Our slope came out to be 1.0196 kg, a little higher than the .950kg we used in our experiment. The percent error that we calculated was 7.3%. This is a fairly good percent error, yet obviously something still went wrong. As shown in question 2, some error in our results came from ignoring friction. Friction would be subtracted from the force, therefore making acceleration a smaller number. In question 4, we examined the mass of the pulley as a source of error. The mass of the pulley would be added to the total mass. Since this number is on the bottom of a fraction it would reduce the value of acceleration. When combining both of these factors together, it is probable that they accounted for a large portion of our error. Both of these changes would change the graph to decrease the slope and make it closer to .950kg, the mass that we actually used in the experiment. In order to get more accurate results in the future, it is important to take both friction and the mass of the pulley into account. Together, these values can make a difference in the results of the produced equation.

= ** Lab: Newton's Second Law of Motion ** = = ** Members: Sam Fihma, Erica Levine, Evan Bloom ** = = ** Period 2 ** = = ** Due Date: December 1, 2010 ** =

**Purpose:**
The purpose of this lab is to show how force, acceleration and mass are all related.

If the force acting upon a system and the acceleration of the mass of the system are compared using the base equation F = m*a, then one should be able to tell that force and acceleration are directly related, resulting in a linear graph in which the slope represents the mass.
 * Hypothesis:**

If the acceleration of a mass and the mass itself are compared using the base equation F = m*a, then the results should show that mass and acceleration are inversely related, with the best-fitting graph being a power graph.

**Procedure:**
Setup: Part 1: Test Relationship Between Acceleration and Force Part 2: Test Relationship Between Mass and Acceleration This picture shows an example of how we extracted acceleration from Data Studio. We found the value for acceleration to use on our graphs by taking the slope of our velocity time graphs. For example, in the situation illustrated by this picture, the slope would be .273 m/s^2.
 * 1) Clamp pulley to edge of a table
 * 2) Place a track on the table, lined up with the pulley
 * 3) Measure the mass of the cart so it can be taken into consideration for calculations
 * 4) Tie one end of a string to the front of the cart, and the other end to the hanging mass
 * 5) Place the cart on the ramp and the hanging mass below the edge of the table
 * 6) Make sure the string runs through the wheel of the pulley
 * 7) Plug the photogate timer into both your computer and the pulley
 * 8) Load Data Studio on your computer
 * 1) Pull the cart back and let go, the hanging mass should accelerate the cart (start recording on data studio)
 * 2) Make sure to stop the cart before it collides with the pulley (stop recording on data studio)
 * 3) Perform multiple trials (steps a-b) to acquire the most accurate data
 * 4) Highlight the slope of the velocity/time graph on Data Studio
 * 5) Under fit, select linear and a linear fit box should appear
 * 6) Record the value of the slope shown in the linear fit box (this is shown below)
 * 7) Repeat Steps (a-f) for varying masses of the hanging mass and mass on the cart
 * 8) Make sure the total mass stays constant. You can do so by moving masses on the cart to the hanging mass.
 * 1) Set the hanging mass to .03 kg and keep this constant throughout all trials
 * 2) Start with a mass of 2,000 g on top of the cart
 * 3) Repeat steps (a-f) from Part 1
 * 4) Repeat steps above (a-c) for varying masses placed on the cart
 * Data and Graphs:**

This is the data and chart where we validate that force is directly proportional to Acceleration.

Sample Calculation to find force: Force = Total Mass x Acceleration Force = 0.529kg x 0.035m/s^2 Force =.019N




 * Analysis Question 1a:** This force v. acceleration graph is linear. We used the different acceleration and force numbers from the chart above. This graph has a good fitted trendline with an equation of y=0.529x and a R squared value of 1. The slope of this trendline is 0.529. This corresponds to the mass (in kg) used in our experiment. The slope and our total mass are the same, which is a percent error of 0 (this is further proved below). Friction has causes some error in this department, but when we conducted our experiment, friction was such a small factor that it did not even make a noticeable effect on our numbers. The slope is equal to the mass because of the equation F=ma. In our graph, force is the y value and acceleration is the x value. Slope is the number that is multiplied by x and in this case, mass is being multiplied by x. With mass as our constant, this shows how force and acceleration are directly proportional. As force is increased, acceleration does too. As force is decreased, acceleration does too. The y-intercept value of this graph is 0, meaning when you plug in 0 for x (no acceleration), y=0 (no force). This means that when there is no acceleration, there is no force moving the object and vice versa. This makes sense because if nothing is acting on a still object, it will not move (Newton's First Law - An object at rest will stay at rest unless acted upon by an unbalanced force.)



This is the data and graph where we validate that acceleration and mass are inversely proportional.
 * Analysis Question 1b:** The graph of Acceleration vs. Mass shown above is not linear. The power of x shown on our graph is -1.087. The power of x should be exactly -1 because in the equation F=ma, f/m represents an inverse. The power shown on our graph is off by less than .1, which tells us our experiment is fairly accurate. The coefficient, or A value in the equation of the line represents the force cause by hanging mass in the experiment. This value on our graph is shown as 0.2236 N. Theoretically, this value should be equivalent to the mass that we placed hanging when we conducted our experiment times gravity (9.8m/s2). The theoretical value is .294 N. The percent error between the two values is 23.9. Although the values are fairly close, there appears to be some excessive error. The variable y represents acceleration in the equation of the line, while the variable x represents mass. This is a different way of saying A= f/m, so the two values should be relatively close.



Sample Calculation to find force: Force = Total Mass x Acceleration Force = 2.529kg x 0.084m/s^2 Force =.211N

The charts and graphs can be found here:

In the first graph, our slope was perfect. However, in the second graph, our slope was a too small. Friction was definitely a source of error in this experiment, especially in the second part of this lab. When we used more mass, friction played a larger role in being a source of error.
 * Analysis Question 2:** For this experiment, friction would cause the acceleration to be lower than it normally would without friction. This is because friction acts against motion and causes the cart to not speed up as much as it would in the absence of friction. To create the same acceleration, there would need to be a greater force. This is due to the fact that force and acceleration are directly related. So if your force increases, so will your acceleration.

The purpose of our experiment was to validate Newton’s Second Law Equation. We did so by testing the relationships between Force and Acceleration and Mass and Acceleration. After constructing a graph of Force vs. Acceleration from collected data, we concluded that the relationship between force and acceleration is direct. We can assume this because of the linear graph that was produced (y= .592x). This graph shows that as acceleration increases, force does too. This makes sense with the equation F=ma, because as the as the a variable increases, the F variable will increase too. From the graph of Acceleration vs. Mass, we concluded that the relationship between the two variables is inverse. The equation of this line is y= .2236x^-1.087. The exponent in this equation is -1.087 which is very close to -1. The exponent -1 represents an inverse, so we can assume that acceleration and mass are inversely proportional. The graph shows that as acceleration decreases, mass increases. This makes sense with the equation F=ma. Assuming force remains constant, as a decreases, m must increase to produce the same F value. The relationships that we observed prove that the equation F=ma is accurate. The percent error for the first graph was 0%. This is because the slope of the graph ended up being exactly the same value as the total mass that we experimented with. The percent error for our second graph was 23.9%. The error is displayed by some outlier points around the line of best fit. The exponent of -1.087 should also be -1, which also displays our error. It is possible that some error stemmed for inconsistency in equipment. On the first day of gathering data, we used a cart that weighed 499 g. On the second day we used a cart that weighed only 493 g. Although we added 6 g of weights to the cart for all of our trials on the second day, it is a possibility that this disparity cause some error. In addition, there may have been trials where the hanging masses fell off which could have changed our results. To improve this experiment in the future, it is important to use exactly the same equipment each date of testing. Also, it is important to be more cautious when conducting trials, and make sure that all weights remain intact throughout the experiment.
 * Conclusion:**

_ = **Lab: Inertial Mass** = = **Members: Sam Fihma, Erica Levine, Evan Bloom** = = **Period 2** = = **Due Date: November 23, 2010** =

Find the mass of a Rubik's Cube based on the time of one period of vibration of other objects with known masses.
 * Objective:**

1. Inertial Balance 2. Clamp 3. Stopwatch 4. Varying masses 5. Rubik's Cube 6. Paper towel (optional)
 * Materials:**

1. Clamp inertial balance to a table. 2. Place a mass on the balance. Use paper towel to avoid any movement inside the balance. 3. Using a stop watch, record the time of 10 periods of vibration. 4. Divide your time by ten in order to get the time of one period. 5. Repeat steps 3 and 4 two more times with the same mass and calculate the average of your three answers. 6. Repeat steps 2-5 with other masses. Recommended masses are: 500, 400, 300, 200, 100, 75, and 50. 7. Create a Vibration vs. Mass chart with your data. Find the line of best fit and its equation. 8. Repeat steps 2-5 with a Rubik's Cube as your mass. 9. Plug in your answer from step 8 into the equation of the line of best fit. Solve for x. This will be the predicted mass of the Rubik's Cube.
 * Procedure:**

Note: Be sure to record all data in a chart.


 * Data Tables:**

Mass vs. Vibration- Multiple Trials

We took the time of the Rubik's Cube 4 times just to be extra sure.

Mass vs. Vibration- Average

This is the average time per period of each mass.


 * Graph:**

The charts and graph can be found here:

Using the equation of the line of best fit in the graph above, we predicted what the mass of the Rubik's Cube to be. The Rubik's Cube had a period of 0.432 seconds, so we plugged this number as y into the equation. After doing the math, we found the predicted mass of the Rubik's Cube to 92.09 grams.

y = 0.0011x + 0.3307 0.432 = 0.0011x + 0.3307 0.1013 = 0.0011x x = 92.09
 * Calculations:**

At first, our percent error was near 17%. We looked at our numbers and realized that the rubik's cube time was very close to our 100g mass time. This shows how we did not make a systematic error, but rather made errors for individual trials. We looked at our chart and saw that 200g, 300g, and 400g masses were not very close to the line of best fit. We redid these values and got a new equation with a better r squared value. With these new numbers, our percent error was down to 9%.
 * Percent Error and Difference:**

Here is the math behind our percent error and difference:


 * Discussion Questions**
 * 1) Did gravitation play any part in this operation? Was this measurement process completely unrelated to the “weight” of the object?
 * 2) Gravity did not play any part in this lab. However, by calculating the unknown mass from the inertial balance, it is possible to find the weight of the object by multiplying the mass by the gravitational force.
 * 3) Did an increase in mass lengthen or shorten the period?
 * 4) The increase in mass lengthened the period. As the mass increased, inertia increased as well.
 * 5) How do the accelerations of different masses compare when the platform is pulled aside and released?
 * 6) The heavier the weight is, the slower the acceleration is. The lighter the weight is, the fast the acceleration is.
 * 7) Did you imagine that the period would have been different if the side arms were stiffer? Would this change, lengthen or shorten, the period of the motion?
 * 8) Stiffer arms would affect the period of the masses. Stiffer arms would shorten the length of the period of motion because the arms would not bend as far, and therefore would complete a period in a shorter amount of time.
 * 9) Is there any relationship between inertial and gravitational mass of the object?
 * 10) Inertial and gravitational mass are directly related. The greater the gravitational mass of an object, the greater the inertial mass of the object is. The two different types of masses are identical except for the method in which to find them.
 * 11) Why do we almost always use gravitation instead of inertia as a means of measuring mass of an object?
 * 12) Using gravity instead of inertia is much easier. Scales are much easier to use than inertial balances. It is also much more accurate and eliminates human error.
 * 13) How would the results of this experiment be changed if you did this experiment on the moon?
 * 14) The results of this experiment would not change if it was conducted on the moon. Gravity does not factor into measuring inertial mass.


 * Conclusion**

The purpose of this experiment was to predict the mass of a Rubik's cube based on vibration periods of known masses. We collected data and were able to create a graph that compared mass to time of 1 period of vibration. Using a best fit line and our value for time of 1 period of vibration of the Rubik's cube, we were able to predict the mass of the Rubik's Cube. Our method of solving this problem proved to be effective, although our prediction was slightly off due to small errors. We estimated the mass of the Rubik's cube to be 92.09 grams, when in reality it was 101.41 grams. We calculated our percent error to be 9.19%. We concluded that this error came from errors in our collected data. It is possible that we started the timer too early or too late, and it is also possible that we miscounted the periods. This would cause the equation of our line of best fit to be altered, therefore producing a different mass for the plugged in time of 1 period of the Rubik's cube. This error in our data most likely occurred because the periods at times moved fast and were challenging to count precisely and accurately. In the future, it is important to collect data as carefully as possible. Using mostly heavier masses would help this problem because the periods wouldn't be too fast to count. The performers of the lab also need to make sure they start the timer at precisely the moment the weight is released. From collecting data in this lab, we were able to conclude that mass and inertia are directly related. The more mass an object has, the more inertia it will have. This is justified by the positive slope of our linear graph shown above.