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course Phy 201
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.
6 count per 1 second. 25 big paper clips and 5 small paper clips were attached at the start. All went 30 cm.
Trial1 – 4 big, 1 small at 7.6s
Trial2 – 6 big, 1 small at 3.5s
Trial3 – 6 big, 2 small at 2.6s
Trail4 – 7 big, 2 small at 1.5s
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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.
Trial 1 – 30 cm / 7.6s = 4 cm/s
Trial 2 – 30 cm / 3.5s = 9 cm/s
Trial 3 – 30 cm / 2.6s = 14 cm/s
Trial 4 – 30 cm / 1.5s = 19 cm/s
those are average velocities, which are useful, but aren't 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.
The line is straight and slope is 15.5
the slope has units, and is calculated from a rise and a run between two selected points on the line. The rise and the run each has units and should be given to show how you get the slope.
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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:
1)3 cm, 23 cm, 13 count
2) 2 cm, 26 cm, 16 ct
3)3 cm, 20 cm, 13 count
The count is a great addition, but it isn't a position. You were asked for three positions.
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Give a brief explanation of what the data mean, including a statement of the units of the numbers:
Average velocity can be solved from the distance and time we found
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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.
trial1 – 3 cm of separation, and it came to rest at 26 cm, taking 2.1s
Trial2 – 2 cm of separation, and it came to rest at 28 cm, taking 2.6s
Trial3 – 3 cm of separation, and it came to rest at 23 cm, taking 2.1s
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Sketch a graph of distance traveled vs. initial separation. Describe your graph.
The less the initial separation the further it traveled. The graph is relatively curved.
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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?
* 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.
coasting distance is 4 cm
Slope would be 2
initial separation divided by coasting distance
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Repeat for the point where initial separation is 5 cm.
coasting distance is 11 cm
Slope is 2.2
initial separation divided by coasting distance
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According to your graph, if the initial separation is doubled, what happens to the distance the car travels?
it decreases
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According to your graph, if the initial separation is doubled, what happens to the slope of the graph?
the slope decreases
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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.
Friction acts negatively, constantly slowing the acceleration of the car. The velocity and magnet force acts positively but friction overpowers both so the car eventually stops.
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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.
20 newtons(?)
I doubt im doing it right.
you might have lost the thread by this point; we'll pick up on this in class
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Repeat for initial separation 5 cm.
12.5 N
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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?
A 7.5 N difference
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How much work was therefore done by the magnetic force between the 5 cm and 8 cm separation?
7.5N
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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? A 3cm displacement
How can you calculate the average force given the displacement and the work? Force divided by displacement
What therefore was the average force? Roughly 2 N/cm
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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? 2Newtons per centimeter
What aspect of the graph of work vs. separation is associated with this average rate? Both axes are used to find the average rate
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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? 4 newtons/cm
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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? Divide work by separation to see how much work would be used at a given point, on average.
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See my notes. We'll be going over this in class, after which you should submit a revision.
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