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course PHY 231
9/18 12:58 pm
As usual these questions do get more challenging toward the end, and it is not expected that everyone will be able to answer them all. Do your best but don't let yourself get too bogged down.Note: numbers as opposed to quantities
* Often a question will ask you to list numbers. When asked for numbers you should give only the numbers, without the units and without any words of description. Of course the units and descriptions are important., so after listing the numbers for all your data you will then be expected to state the units and the meaning of your numbers in a subsequent line.
* If a question asks for quantities, then the units should be included. A brief explanation of meaning is generally expected after you complete your listing of quantities.
For the car and paperclips
Let 1 unit of force correspond to the weight of 1 small paperclip. On this scale the weight of a large paperclip is 4 units of force.
If the paperclip is on the car, its weight is balanced by the upward force exerted by the table and it has no direct effect on the car's acceleration. If it is suspended, then its weight contributes to the accelerating force.
You should have obtained a count and a distance from rest for each trial, and on each trial there will be some number of force units suspended from the thread (1 unit for every small, 3 units for every large clip). Report in the first line the number of clips, the count and the distance from rest, separated by commas, for your first trial. Report subsequent trials in subsequent lines. After reporting the data for all your trials, give a brief explanation of your setup and how the trials were conducted. Include also the information about how many of your counts take how many seconds.
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11, 7, 9
12, 6, 16
13, 7, 23
17, 7, 70
20, 4, 70
22, 3, 70
For this I would set the car down at my starting point with the clips hanging over the table, then let it go and start counting. Now, when I started looking at my numbers I realized I had a disparity. My first three trials the car came to a stop while the last three stopped only because they ran out of room, meaning that for the former, vf=0, and for the latter, vf≠0. I consider the second set to be the ones I need to use from here on out but I will give some of the values asked for for both sets, the first in italics. I'm not certain about this though but my graph really bares out that there is a difference between these two sets. My counts are 3.93 second per one 8 count.
If the car came to rest, then vf = 0 for that trial and the acceleration would be in the direction opposite velocity (i.e., the negative direction, if you declare the direction of motion to be positive).
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Determine the acceleration for each trial. You may use the 'count' as your unit of time, or if you prefer you can convert your counts to seconds and use seconds as your time unit. In each line below list the number of suspended force units and your acceleration. After reporting your results, give in the next line the units and your explanation of how your results were obtained.
19, 0.51
20, 0.889
23, 0.923
31, 2.86
40, 8.75
47, 15.56
!!!!!! Above you gave two different force units for the big clips, 3 and 4. I have used the first quantity, 3, for my calculations.
typo on my part; you chose correctly
My units are in cm/count^2.
The first three are the vf=0 trials and you can see that they have a fractional value for acceleration whereas the second set has a much larger acceleration value associated with it.
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Sketch a graph of acceleration vs. number of force units, and describe your graph. Fit a straight line to your graph and determine its slope. Describe how the trend of your data either indicates a good straight-line fit, or how it deviates from a straight line.
Okay this is where my two sets of data really show just how different they are. My first set fits to a very different line than the second. It slope is very small, 0.1, whereas the second sets slope is 7/10. They data points do fit well to their corresponding lines though with a relatively small amount of deviation.
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For the toy car and magnets:
Give the three positions you measured for each trial, one trial to each line. Each line should consists of three numbers, representing the position in cm of the fixed magnet, the position in cm of the magnet of the toy car, and the position in cm at which the car came to rest after being released. This is your raw data:
3, 14, 16
3, 13, 16.5
3, 12, 18
3, 11, 19.5
3, 10, 22
3, 9, 24
3, 8, 28
3, 7.5, 32
3, 7, 34
3,7, 45
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Give a brief explanation of what the data mean, including a statement of the units of the numbers:
These numbers are all based on a meter stick so that the fixed magnet was always at 3 cm from one end and then the initial position moved closer and closer to the limit of 6 cm beyond which the car flipped around and stick to the magnet the last number is where the car came to a stop with respect to the meter stick. All of these values are in cm.
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For each trial, give the distance of separation between the two magnets at the instant of release, and the distance the car traveled between release and coming to rest. Give in the form of two numbers to a line, separated by commas, with separation first and coasting distance second. After the last line, give a brief explanation of how your results were obtained and what the numbers mean, including a statement of the units of the numbers.
11,2
10, 2.5
9, 6
8, 8.5
7, 12
6, 15,
5, 20
4.5, 24.5
4, 27
3, 39
Here I just subtracted my fixed magnet position from the initial position of the car magnet to find the amount of distance between them. I did the same for the displacement by subtracting the initial position from the final position to get the distance travelled by the car. The units here are still just cm.
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Sketch a graph of distance traveled vs. initial separation. Describe your graph.
My graph is starts high for distance travelled and move toward 0 as initial position becomes larger. It is a sloping curve that is concave up.
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Sketch the smooth curve you think best represents the actual behavior of distance traveled vs. initial separation for this system.
This is, like above, a sloping, concave up, curve. The points fall within a reasonable amount of deviation from the curve.
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Identify the point where initial separation is 8 cm.
* What coasting distance corresponds to this point?
* What is the slope of your smooth curve in the neighborhood of this point?
Give the coasting distance as a number in the first line, the slope of the graph as a number in the second line. Starting in the third line give the units of your quantities and explain what each quantity means, and how you obtained it.
8.25, -3.5
The first number is the displacement in cm and the second number is cm/cm, i.e. -35 cm/ 10 cm. The first is just how far the car travelled, while the second is how much distance the car is moving per initial position, i.e. -3.5 cm distance for every +1 cm initial position, and is thus a rate of change. I obtained this by taking the initial position values around 8, I used 7.5 and 8.5, and divided this by the change is distance travelled of those two positions, -3.5 cm/ 1 cm, so -3.5 cm less distance travelled for 1 change in initial position.
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Repeat for the point where initial separation is 5 cm.
21, -6.25
In cm and cm/cm respectively. Found just like the above.
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According to your graph, if the initial separation is doubled, what happens to the distance the car travels?
The second separation has a fractional distance as compared to the first, such as 3 and 6, there is very large change from the first's distance to the second's
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According to your graph, if the initial separation is doubled, what happens to the slope of the graph?
It decreases greatly at first, such as 6 and 3, but then less and less so, say 5.5 and 11. The slope of the line running between these two points becomes less and less from the first set to the second set.
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The distance the car travels is an indication of the energy it gained from the proximity of the magnets. Specifically, the frictional force slowing the car typically is about .01 Newton, or 10 milliNewtons. The force exerted on the car by friction is in the direction opposite the car's displacement, so when you calculate the work done by this force, your force and the displacement will have opposite signs (i.e., one will be positive and the other will be negative).
Using the .01 Newton = 10 milliNewton force and your displacement in meters (you likely calculated the displacement in centimeters, so be sure you use the equivalent displacement in meters) find the work done by this force on each of your trials. Give below the initial separation of the magnets, the work done by the frictional force acting on the car in Newtons, the work in milliNewtons, in the form of three numbers per line separated by commas. In the first subsequent line, explain your results and include a detailed sample calculation.
3, -0.0039, -0.39
4, -0.027, -0.27
5, -.002, -0.2
6, -0.0015, -0.15
7, -0.0012, -0.12
8, -0.00085, -0.085
9, -0.0006, -0.06
10, -0.00025, -0.025
11, -0.0002, -0.02
This is the work done for each initial position and the work is negative because I had my distance as positive but friction was working against my car so it was pushing backward away from the direction the car was travelling. Here I took my distances travelled and divided by 100 to obtain meters and then I multiplied this by the force of the friction, using 0.01 N. I based this on the equation W= F*d where my force was 0.01 N and my distance was how far the car had moved for that initial position. My units are first cm, then N*m/J, and last mN*m, or mJ. The last value was found by multiplying the N*m work by 100 to get mN*m.
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Friction does negative work on the coasting car, which progressively depletes its kinetic energy (recall that kinetic energy is energy of motion). In this situation the original kinetic energy of the car came from the configuration of the magnets (the closer the magnets, the greater the KE gained by the car). We say that the initial magnet configuration had potential energy, with closer magnets associated with more potential energy. The initial potential energy of the magnets was therefore converted into the initial kinetic energy of the car, which was then lost to friction as friction did work on the car equal and opposite to its initial kinetic energy. (actually the potential-to-kinetic-energy transition takes place over a significant interval of time and distance, so it would be more appropriate to speak of the work by friction on the car begin equal and opposite to the initial potential energy, but we won't really worry much about that just yet).
Explain this below in your own words, as best you understand it.
The car gained potential energy based on how close it was to the fixed magnet because of the repulsive electromagnetic force between the two magnets. When the car is released this turns to kinetic energy. The kinetic energy is expended by the car overcoming the force of friction, which is pushing in the opposite direction of the car's motion.
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... impulse ...
The car also exerts a frictional force on the table which is equal and opposite the frictional force exerted on it (if the table was on frictionless rollers the frictional force exerted by the car would cause the table to accelerated very slowly in the direction of the car's motion), so it does work against friction which is equal and opposite to the work friction does on it. So the car does positive work against friction. This work is done at the expense of its kinetic energy.
If you were to make a table showing work done by the car against friction vs. initial separation it would be the same as the table you gave previously, except that the work would be positive (you did remember to make the work negative on the previous table, didn't you?). I'm not going to ask you to give that table here, since except for the sign of the work it is the same as your previous table.
What we are going to want is a graph of the work done by the car against friction, vs. initial separation.
... to cheerios ... add to table fraction of cheerio, then correct for my 15% efficiency
??????? Not sure what this means or how to do this either.????????
Now you already have a graph of distance vs. initial separation. You can add a new labeling to your vertical scale to represent the corresponding work done by the car against the frictional force. If you don't understand what this means, you can go ahead and create a separate graph of work done vs. initial separation. Either way:
According to your new graph, or the new labeling of your original graph, what is the work done by the car against friction when the initial separation is 8 cm? Give the quantity in the first line, a brief explanation in the second.
0.085 mN*m or mJ
Well, conveniently enough, all I have to do is divide my graph value by 100 to obtain the mN*m value of the work done.
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Repeat for initial separation 5 cm.
0.2 mN*m
Found just as above.
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According to your graph, how much more work was done against friction when the initial separation was 5 cm, than was done when initial separation was 8 cm?
0.2-0.085= 0.115 mN*m
I think this is what you're asking for. I am not completely certain about any of the following answers. The difficulty is not knowing exactly what it is the question is asking for but I have made by best attempts to follow along.
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How much work was therefore done by the magnetic force between the 5 cm and 8 cm separation?
0.428 mN*m
For this I think the work done is represented by the area of the graph between 8 cm and 5 cm so that is how I obtained my value.
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For a system released at a separation of 5 cm or less, the magnets exerted a decreasing force between the 5 cm separation and the 8 cm separation. The force has an average somewhere between the forces exerted at the 5 cm and 8 cm separations. If you answered the preceding question correctly you know the work between the two positions.
Through what displacement did the magnetic force act between these two separations?
How can you calculate the average force given the displacement and the work?
What therefore was the average force?
0.115 m
Here I think the first part is found by subtracting the distance at 5 cm from the distance at 8 cm, so that is what I have given.
Well we have W_Ave and ∆d so we use W_Ave=F_Ave*∆d to get F_Ave = W_Ave/∆d. This gives F_Ave= 1.24 mN. I think I understand the process here but am not certain that I have used the correct quantities to find this.
You wouldn't average the values for work; you found above that .115 mN * m of work was done between the two points. You didn't give the values that gave you your result for F_ave, but it would be F_ave = .115 mN * m / (3 cm) = 3 mN or so.
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What therefore is the average rate at which work is being done by the magnets, per unit of separation, between the 5 cm and 8 cm separations?
What aspect of the graph of work vs. separation is associated with this average rate?
I'm not sure about this but I think this is the change in work vs change in position, which is (0.085 - 0.2) / (8 -5) = -0.038 mN*m / cm. Based on my method here this represents the approximate slope of the graph on this interval.
m / cm = 100, so this is -.038 mN * 100 = -3.8 mN.
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What is your best estimate of the average rate at which work is being done by the magnets, per unit of separation, when the magnets are 6 cm apart?
Using my values above, which clearly may be way off, I get approximately 0.162 nN*m, which is somewhat close to my experimental value of 0.15.
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In general how can you use a graph of work vs. separation to this system to find the force exerted by the magnets at a given separation?
The force should be slope of the graph at the given separation. I know this is true but don't know if my answers above actually reflect that.
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You are right about the slope being the force; how does this relate to your answers to the last few questions? No need to submit an answer, just think about it and when the light bulb comes on (as I think it will), drop me a quick note using the Submit Work Form.
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