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course phy 201
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 susequent 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.
13,30cm,
19,30cm,
21,30cm,
23,30cm,
24,30cm,
32counts
20counts
18counts
14counts
11counts
Transfer clips from magnet to end of thread. The car had 25 big clips and 5 small ones and one big one on the string. We moved the car 30cm each trial. The timer was count 1 thru 8 single times so 32 counts means 4, 8 counts.
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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.
30/32=.93cm/counts
30/20=1.5cm/counts
30/18=1.6cm/counts
30/14=2.14cm/counts
30/11=2.72cm/counts
Final position-Initial position which gave me 30cm/time
That's not the acceleration, it's the average velocity. It's the first step in reasoning out the acceleration.
You should reason out the accelerations.
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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.
Unit force
.8
1.07
1.36
Well each graph tends to increase, and the slope is positive.
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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:
0cm,12cm, 18.5cm
0cm,8cm,25cm
0cm,6cm,33cm
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Give a brief explanation of what the data mean, including a statement of the units of the numbers:
The magnet was set a 0cm the car was being release at the second number and the third number is the distance it traveled. The measurements was 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.
12,6.5
8,17
6,7 do you mean 6, 27?
Well the first number is the released point which is the same because I stated at a fixed point of 0cm. Then the second number I just subtracted the released point by the stop position.
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Sketch a graph of distance traveled vs. initial separation. Describe your graph.
They both have the same starting point, but the distance traveled has a positive incline, the initial separation does not a have steady increase.
You appear to be describing two different graphs. There is only one distance vs. init separation graph, and it contains points (6.5, 12), (17, 8) and (27, 6).
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Sketch the smooth curve you think best represents the actual behavior of distance traveled vs. initial separation for this system.
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Identify the point where initial separation is 8 cm.
* What coasting distance corresponds to this point?
17cm
* What is the slope of your smooth curve in the neighborhood of this point?
21cm
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.
Because of the curve I came up on the graph
I don't think you're reporting the slope; can you explain how you got the slope?
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Repeat for the point where initial separation is 5 cm.
39cm
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According to your graph, if the initial separation is doubled, what happens to the distance the car travels?
If the initial separation double the distance would decrease.
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According to your graph, if the initial separation is doubled, what happens to the slope of the graph?
The slope would decrease as well.
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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.
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.
Well friction slows the car down which then also slows down the kinetic energy. The closer the magnets the greater the KE gained by the car. We first started out with potential energy and once the car left it converted into KE which lost because do to friction.
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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
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.
25cm
25 cm isn't work. Can you insert the requested explanation, along with a calculation of work = force * distance?
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Repeat for initial separation 5 cm.
40cm
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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?
Difference between 25 and 40
15cm
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How much work was therefore done by the magnetic force between the 5 cm and 8 cm separation?
8-5cm=
3cm
That is the displacement, not the work.
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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?
15cm
How can you calculate the average force given the displacement and the work?
Fnet=ma
What therefore was the average force?
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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?
14cm
What aspect of the graph of work vs. separation is associated with this average rate?
14/3
4.66
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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?
33cm
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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?
1 newton is the net force required to accelerated a kg mass at 1 m/s
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Most of your data appear to be good. You've answered some questions correctly, others you haven't. Check out my notes.
You should submit a revision for most of the noted items.
&#Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
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