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course Phy 241
9/20 2!!!!!!!!!!This is not the complete assignment, but I wanted you to know I have been working on it.!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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.
2Sm & 1Lrg, 1sec, 1.5cm
3Sm & 1Lrg, 1sec, 1.5cm
4Sm & 1Lrg, 5sec, 23.5cm
5Sm & 1Lrg, 6sec, 50cm
1Sm & 2Lrg, 2sec, 7.5cm
2Sm & 2Lrg, 4sec, 41cm
2Sm & 2Lrg, 5sec, 50cm!!!!Repeat of previous line, diff data collected!!!!
2Sm & 3Lrg, 3sec, 50cm
3Sm & 3Lrg, 2(3/4)sec, 50cm
3Sm & 3Lrg, 2sec, 50cm!!!!Repeat of previous line, diff data collected!!!!
4Sm & 3Lrg, 1(3/4)sec, 50cm
5Sm & 3Lrg, 1(1/3)sec, 50cm
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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.
force =5, aAve = 3
force =6, aAve = 3
force =7, aAve = 1.88
force =8, aAve = 2.7778
force =7, aAve = 3.75
force =8, aAve = 5.125
force =8, aAve = 4
force =11, aAve = 11.1111
force =12, aAve = 13.2231
force =12, aAve = 25
force =13, aAve = 32.6531
force =14, aAve = 56.25
Force was found by substituting values of 1 for sm paper clip and 3 for lrg paper clips and adding the values together.
`ds(displacement)/`dt = vAve
vAve*2 = vf (while v0 = 0)
vf/`dt = aAve (while v0 = 0)
?????There is some question about 1 Lrg = 3 units, because for 2 lrg & 1 sm the aAve was greater than 1 lrg and 5 sm. Seems to be a very good approx. and I’m sure there are other lurking variables involved in our experiment?????
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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.
???I did not get to gather as much data as I would have liked, so for the first several experiments my data is not very consistent.?????
From the point (5, 3) to (14, 56.25), slope equals rise/run = 53.25/9
????I calculated slope from these two points, because I could not fix a line very well with my data?????
My graph appears to be exponential, which would mean a straight line fit would not be appropriate. Which means it appeared to be increasing at an increasing rate.
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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, 7, 41.5
1, 8, 34
1, 9, 28
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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 always at the fixed position of 1cm. When the car was forced to the 7 cm mark(which was actually 6cm from the magnet) the resulting force caused the car to travel 41.5 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.
6, 41.5
7, 34
8, 28
Results were obtained by subtracting 1 from starting position of the car due to the fact that the magnet was always at the 1 cm mark. So the first number is distance between 2 magnets at beginning of experiment in cm. Second number is distance traveled by the car in cm
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Sketch a graph of distance traveled vs. initial separation. Describe your graph.
Graph appears to be concave up or decreasing at a decreasing rate.
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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.
28
-6
For initial separation is 8 cm the car travels 28 cm which was obtained because happened to be part of my data collected. Slope = rise /run, where `dy = -6cm(diff of dist travled between intial starting points 7cm&8cm) and `dx = 1( 8cm – 7cm). So slope is approx. -6/1 or -6cm for every cm moved away from the magnet not on top of the car
very reasonable estimate; can you translate that into terms of energy? That wasn't asked, just something to think about.
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Repeat for the point where initial separation is 5 cm.
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According to your graph, if the initial separation is doubled, what happens to the distance the car travels?
It appears that the car travels between 2 and 3 times as far for smaller dist between the car and the fixed magnet.
Ex) For initial sep of 5 cm car travled 62cm, for initial sep of 10 cm car travled 24.5cm
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&#Good responses. See my notes and let me know if you have questions. &#
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course Phy 241
9/20 2!!!!!!!!!!!!!This again is only a partial assignment!!!!!!!!!!!!!!!!!
Brief at-home experiment
Is the magnitude of a ball's acceleration up an incline the same as the magnitude of its acceleration down the incline?
Test this.
One possible suggested method: Use the short ramp to get the ball started, and let it roll from the short ramp onto the longer ramp, with the longer ramp inclined so the ball rolls up, rather than down. Get the ball started on the short ramp, either by inclining it toward the long ramp or giving it a push (before it reaches the long ramp). Click the TIMER at the instant the ball hits the 'bump' between the two ramps, again when the ball comes to rest for an instant before accelerating back down the long ramp, and once more when the ball again hits the 'bump'.
Everyone will tend to anticipate their 'clicks', and to try to compensate for their anticipation. If the effect of anticipation is the same for all three timed events, then uncertainties in your results will all be due to the TIMER. If the degree of anticipation differs, as it inevitably will, then the uncertainties are compounded. If the degree of anticipation (and/or compensation) tends to be either greater or less for the second event (the ball stopping for an instant) than for the first and third (the 'clicks' made when the ball hits the 'bump'), then a systematic error is introduced (you will have a tendency to 'short' one interval while extending the other).
If there exists a difference between the times up and down the ramp, and if the difference is great enough to show through the uncertainties and systematic errors, then you might get a useful result (e.g., either the actual times are the same, or they differ).
Give a brief report of your data and your results:
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I do not have the equipment to do this, will get it done after equipment is acquired
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Displacement and force vector for a rubber band
The points (2, 1) and (4, 6) have the following characteristics:
* From the first point to the second the 'run' is 4 - 2 = 2 and the 'rise' is 6 - 1 = 5.
* The slope of the line segment connecting the points is therefore 5 / 2 = 2.5.
* A right triangle can be constructed whose hypotenuse is the line segment from (2, 1) to (4, 6), and whose legs are parallel to the coordinate axes. The leg parallel to the x axis has length 2, and the leg parallel to the y axis has length 5. The hypotenuse therefore has length sqrt(2^2 + 5^2) = sqrt(29) = 5.4, approx..
* The area of the trapezoid formed by projecting the two points onto the x axis, whose sides are the projection lines, the segment [2, 4] of the x axis and the line segment between the two points, has 'graph altitudes' 1 and 5, giving it average 'graph altitude' (1 + 6) / 2 = 3.5, and width 5 - 1 = 4, so its area is 14.
* The vector from the first point to the second has x component 2 and y component 5. Its length is sqrt(20) and its direction is specified by the ratio of the y to the x component.
Any time you see two points on a graph you should be aware that you can easily construct the right triangle and use it to calculate the slope of the segment joining them, the distance between the points, the area of the associated trapezoid and the components of the vector. Depending on the nature of the graph, some of these quantities will sense and be have useful interpretations, while others probably will not.
In the particular example of a rubber band stretched between the two points, with the coordinates in centimeters, the most important quantities are the components and length of the vector. The vector from the first point to the second is in this case a displacement vector. The displacement vector has x component 2 cm, y component 5 cm and length sqrt( 2 cm)^2 + (5 cm)^2 ) = sqrt( 29 cm^2) = sqrt(29) cm, about 5.4 cm.
* We denote the displacement vector from (2 cm, 1 cm) to (4 cm, 6 cm) as <2 cm, 5 cm>
If we divide the this vector 2 we get a vector of length 2.15 cm. The x component will be 1 cm (half of the original 2 cm) and the y component 5 cm /2 = 2.5 cm (half of the original 5 cm). The ratio of the components is the same, so this vector will be in the same direction as the original vector. If we multiply this vector by 2 we get a vector of length 8.6 cm. Its components will be double those of the original vector, and it will also be in the same direction as the original vector.
If we divide our vector by 5.4 cm, then the resulting vector has length 5.3 cm / (5.3 cm) = 1. The x and y components will be 2 cm / (5.4 cm) = .37 and 5 cm / (5.4 cm) = .92. As before, the ratio of the components is the same as for the original vector (accurate to 2 significant figures),
The vector <.37, .92> has magnitude 1 (calculated to 2 significant figure), as you can easily verify using the Pythagorean Theorem. Its direction is the same as that of the displacement vector <2 cm, 5 cm>.
Now we invoke a rule to find the tension in the rubber band. Assume that for this rubber band the tension is .5 Newtons for each centimeter in excess of its 'barely-exerting-force' length of 4.8 cm.
* The 5.4 cm length of the stretched rubber band is therefore .6 cm in excess of the 'barely-exerting-force' length of 4.8 cm.
* According to the rule, we conclude that the tension is therefore .6 cm * .5 N / cm = .3 N.
The vector <.37, .92> has magnitude 1, and its direction is the same as that of the displacement vector. A rubber band can exert a force only in the direction of its displacement vector.
* If we multiply our vector <.37, .92> , which we again mention has length 1, by .3 N, we will obtain a vector whose magnitude is .3 N. (whatever quantity we multiply by a vector of length 1, we end up with a vector whose length is equal to that quantity, or more correctly whose length represents that quantity)
* The direction of this vector will be the same as that of our original vector.
* <.37, .92> * .3 N = <0.11 N, 0.28 N>
* This is the tension vector. It tells us that the tension of this rubber band has x component 0.11 N and y component 0.28 N.
Using your data from the rubber band experiment, check your previous calculations of the displacement vector, the unit vector and the force vector for each rubber band. If your original work on this experiment was not done correctly, please correct it and resubmit. Please acknowledge that you have read this instruction and understand it.
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Even though I was not fully aware of what I was doing it seems to have been done correctly.
I understand the above material
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Linear force function and work\energy
If the graph below indicates the tension in Newtons vs. length in cm for a rubber band:
At what length does the rubber band begin exerting a force?
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5cm
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What is the average force exerted between the lengths x = 5 cm and x = 7 cm?
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.5 Newton’s
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As the rubber band is stretched from length 5 cm to length 7 cm, through what distance is the force exerted?
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2cm
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How much work is therefore done in stretching the rubber band from 5 cm to 7 cm length?
`dW = Fave*`ds = .5N*2cm
= 1N*cm !!!!1meter = 100cm, therefor
=1N*.01m = .01 Joule
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How much work is done in stretching the rubber band from the 5 cm length to the 8 cm length?
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`dW = Fave*`ds = .75N*3cm
= 2.25N*cm !!!!1meter = 100cm, therefor
=2.25N*.01m = .0225 Joule
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How much work is done in stretching the rubber band from the 6 cm length to the 8 cm length?
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`dW = Fave*`ds = 1N*2cm
= 2N*cm !!!!1meter = 100cm, therefor
=2N*.01m = .02 Joule
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What are the areas beneath the graph between each of the following pairs of lengths:
* x = 5 cm and x = 7 cm, area = 1 newton/cm
* x = 5 cm and x = 8 cm, area = 2.25 newton/cm
* x = 6 cm and x = 8 cm, area = 2 newton/cm
* x = 7 cm and x = 8 cm, area =1.25 newton/cm
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Verify that the equation of the straight line in the given graph is F(x) = (x - 5) * .5, where F(x) is force in Newtons when x is position in centimeters.
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For x =5, F(5) = (5-5)*.5 =0
For x =6, F(6) = (6-5)*.5 = .5
…. etc….
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University Physics Students: Confirm your results for graph areas by integrating the force function over appropriate intervals. Suggested method: Integrate the function symbolically from x = a to x = b, then use the resulting expression for appropriate values of a and b.
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What is an antiderivative of (x - 5) * .5? Let u = x - 5; the new integrand becomes .5 u du, etc..
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Nonlinear force function and work\energy
The graph given previously was linear. That graph would be realistic for a well-made spring, but not for a rubber band.
The graph given below is more realistic.
Estimate the average force exerted by this rubber band between the x = 5 cm and x = 6 cm lengths. Give your estimate in the first line, and explain how you made the estimate in the second:
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.5N
Estimated because while looking at x mid value it appears force fn appears to be approx. .5N
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Repeat, for the x = 6 cm to x = 7 cm length interval.
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.8N
Estimated because while looking at x mid value it appears force fn appears to be approx. .8N
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Repeat, for the x = 5 cm to x = 5.5 cm length interval.
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.333N
Estimated because while looking at x mid value it appears force fn appears to be approx. .3333N
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Repeat, for the x = 5.5 cm to x = 6 cm length interval.
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.6667N
Estimated because while looking at x mid value it appears force fn appears to be approx. ..6667N
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Estimate the work done in stretching the rubber band for each of these intervals. Give your four estimates in the first line below, separated by commas. Starting in the second line explain how you got your estimates:
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.005 Joule, .008 Joule, .001665 Joule,.00333335 Joule
`dW = Fave*`ds, which is force in cm so we then mult by .01 to get force in meters so we know jouels.
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How much work do you estimate is done between the x = 5 cm length and the x = 7 cm length? Give your estimate in the first line, your explanation starting in the second.
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1.3 joules
Because `dW is related to area, we can simply add the two areas together to get 1.3 joules.
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If a trapezoid was constructed by projecting the x = 5 and x = 7 points of the graph, what would be its area, and would this result be an overestimate or an underestimate of the actual area beneath the graph between these two points?
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Area would be Fave*`dt(which would be 2) and it would be an underestimate.
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&#Your work looks good. See my notes. Let me know if you have any questions. &#