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course Phy 231
12/06/2011 @ 6:47p.m.
Magnets and cars
A car with a magnet on the end was brought into proximity with an opposing magnet and released. The distance traveled by the car was observed as a function of proximity.
The car was also timed as it coasted to rest over a variety of distances.
In some experiments the opposing magnet was brought into proximity with the car, which was free to accelerate as soon as the magnets were sufficiently close.
In some experiments the car was permitted to roll down an incline into proximity with a fixed magnet; the release point, distance of closest approach and rebound distance up the ramp were observed
In some variations two cars with opposing magnets were released simultaneously and distances from release point observed.
Insert a copy of your data here, along with any previously submitted work you wish to include:
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Magnet Car Lab 2
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course Phy 231
11/8/2011 @ 2:18 p.m.
My partner and I attempted this lab in two different ways; first, we held a car stationary with a magnet fastened to the back of it and brought a magnet into the desired distance. We then released the car and measured the distance the car traveled. This was done at various distances to measure the amount of PE relative to how far the magnet was from the car. The second method we used to gather data was quickly sliding the magnet to the desired distance away from the car and measuring the distance the car traveled. The first method had potential error due to the possibility for releasing the car poorly due to human error. The second method had potential error due to the fact that it is extremely difficult to slide a magnet in quickly and having it exactly the desired distance away from the car. The second method is probably more reliable, though it is very possible to have up to a centimeter of error in the distance the magnet was away from the car.(It just occurred to me that it is really pointless to say “this trial is going to have the magnet at 5 cm away since we cannot reliably get it 5cm away while sliding the magnet in quickly. We should have just slid it in a random distance and recorded that distance instead of saying this is the 5cm trial when it could have been 4 or 6 cm easily.)
@& Good idea, good thinking.*@
The following data was recorded using the “first” method of gathering data.
The first number is the trial number, the 2nd number is the distance the magnet was from the car in centimeters, the third number is the distance the car traveled in centimeters.
1, .5, 37.8
2, 1, 36.5
3, 1.5, 29.1
4, 2, 23
5, 2.5, 19.9
6, 3, 17.8
7, 3.5, 14.4
8, 4, 12.6
9, 4.5, 10.4
This data suggests that the further the magnet is from the car, the less distance the car will travel. It also shows that this relationship is not quite linear.
The following data was recorded using the “second” method of gathering data.
The first number is the trial number, the 2nd number is the distance the magnet was from the car in centimeters, the third number is the distance the car traveled in centimeters.
1, 1, 29.3
2, 1.5, 25
3, 2, 23.5
4, 2.5, 18.5
5, 3, 15.8
6, 3.5, 9.3
7, 4, 7.4
8, 4.5, 5.6
The relationships shown here are very close to from the first trial, but we did see smaller distances traveled overall. It isn’t extremely clear why this might be other than possible human error (not exact distances away or the magnet approaching at less-than-ideal angles).
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Self-critique (if necessary):
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Self-critique rating:
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@& Consider that the car started moving before the magnet got to its final position. Could this explain the discrepancies you so clearly observed?*@
@& Good work and good thinking overall.*@
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Based on your observations:
As the car coasts across the tabletop:
What is its acceleration?
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ALL OF THE FOLLOWING CALCULATIONS USE THE FIRST TECHNIQUE OF GATHERING STATED ABOVE.
The average time it took the car to stop was .5seconds, the ‘dV was about -59.2cm/sec, which means a = -118cm/sec^2
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How much frictional force is exerted per gram of the car’s mass?
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127g * -118cm/sec^2 = -15,000 dynes.
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This is the frictional force on the 127 gram total mass, not the frictional force per gram.
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How much work is done by friction per centimeter of coasting distance, per gram of the car’s mass?
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W = F*’ds = -15,000 dynes * 14.8cm = -222,000 dynes * cm
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This is the total work done, not the work per cm.
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How much energy is dissipated by friction, per centimeter of coasting distance, per gram of the car’s mass?
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I believe ‘dW = ‘dKE, so -220,000dynes*cm
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This is total energy dissipated, not dissipated energy per gram \.
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Pick one of your trials. How much energy was dissipated by frictional forces for this trial, and what was the initial proximity of the magnets?
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Trial #2: a = -146cm/sec^2, m = 127g, thus Ffrict = -18,542 dynes
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Graph coasting distance vs. proximity, and relabel your graph to show the energy dissipated by friction vs. the initial proximity of the magnets.
Partition your horizontal axis at half-centimeter intervals and find the slopes associated with the corresponding trapezoids.
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-1000, -7600, -6000, -3200, -2200, -3200, -2000, -2200 (All slopes are in dynes/cm)
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Energy would be in dynes * cm, so rise / run would have units of dynes * cm / cm.
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What is the meaning of the slope of a typical trapezoid?
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Amount of energy dissipated due to friction per unit of distance.
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The amount of energy dissipated by friction per unit of distance has the units indicated in my preceding note.
Think a little more about what this might mean.
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The meaning is very important and revealing so be sure you follow up on this question.
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Make a table of trapezoid slope vs. the midpoint of the corresponding proximity interval, and sketch the corresponding graph.
What is the meaning of your graph?
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Average amount of energy dissipation relative to the distance the magnet was from the car.
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Divide your graph into trapezoids at half-cm intervals and determine the area of each.
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I don’t quite understand what you want from this; the graph is slope which is friction per unit of distance vs. midpoint which is a unit of energy dissipated from friction…though me misinterpreting the question or graph is highly likely.
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I agree that the graph shows friction dissipated per unit of distance.
You can divide this graph into half-cm intervals and calculate trapezoid areas.
What would the area of a trapezoid of such a graph represent?
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What is the meaning of the area of a typical trapezoid?
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If you were to measure magnet force vs. magnet proximity you could construct a graph of force vs. proximity. Based on this graph how could you obtain a graph of the potential energy of the magnet system vs. proximity?
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In this system, the main form of energy dissipation is due to friction, PE is equal and opposite to the amount of work done by friction.
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That's correct and relevant, but it doesn't yet answer the question of how you would use a force vs. proximity graph to obtain the PE vs. proximity graph.
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My first three notes are mainly for your information; however be sure you take a couple of minutes to calculate the requested quantities and think about what they mean.
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More importantly think more about the interpretation of the trapezoids in the various questions.
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