#$&*
course Phy 121
11 pm July 5
Copy this document, from this point down, into a word processor or text editor. Follow the instructions, fill in your data and the results of your analysis in the given format.
Regularly save your document to your computer as you work.
When you have completed your work:
Copy the document into a text editor (e.g., Notepad; but NOT into a word processor or html editor, e.g., NOT into Word or FrontPage).
Highlight the contents of the text editor, and copy and paste those contents into the indicated box at the end of this form.
Click the Submit button and save your form confirmation.
Most students report completion time between 2 and 3 hours for this exercise. Some report significantly longer, some as short as 30 minutes. The variation in times might be related to variations in the amount and nature of graphical analysis in prerequisite courses.
Introductory Remarks:
Recall the introductory problem about the skater propelled by the force exerted by a bungee cord.
It should be obvious that if two forces are exerted through a fixed distance, then if the conditions are otherwise the same the greater force will result in the greater coasting distance.
Similarly it makes sense that if the forces are the same, but one is exerted through a greater distance than the other, the force exerted through the greater distance will result in the greater coasting distance.
In fact it turns out twice the force exerted through the same distance will result in twice the coasting distance, and the same force exerted through twice the distance will also double the coasting distance.
The question is then what would happen if twice the force was exerted through twice the distance.
Doubling the force alone would double the coasting distance; is we then double the distance through which the force is exerted this would again double the coasting distance, resulting in four times the coasting distance.
Also note that since vf^2 = v0^2 + 2 a `ds, we have
a `ds = 1/2( vf^2 - v0^2).
The product of acceleration with displacement is equal to half the change in the squared velocity.
So if acceleration a is doubled while `ds remains the same:
This obviously doubles a `ds, so it doubles 1/2 (vf^2 - v0^2). If 1/2 (vf^2 - v0^2) is doubled, then vf^2 - v0^2 is also doubled.
So doubling a `ds doubles vf^2 - v0^2.
If acceleration and `ds are both doubled, then the result is that a `ds is quadrupled--it becomes 4 times as much--so that vf^2 - v0^2 is also quadrupled.
This is a little easier to understand if we consider the special case where v0 = 0.
In this case we have just (a `ds) = 1/2 vf^2.
So doubling a alone would cause vf^2 to double.
And doubling `ds alone would cause vf^2 to double.
So doubling both a and `ds would cause vf^2 to quadruple.
Finally recall that on a graph of v vs. t, the area of a trapezoid is equal to its average 'height' multiplied by its 'width', which corresponds to multiplying the average velocity by the time interval: area = vAve * `dt.
Keep these things in mind as you work through the following.
The figure below depicts force vs. length for a rubber band.
If you have a printer handy you can print out the figure and sketch a smooth curve to represent what you think is the actual behavior of this rubber band. If not you can visualize this curve.
Answer the following questions:
According to your smooth curve now much force was the rubber band exerting when its length was 100 mm?
According to your curve now much force was the rubber band exerting when its length was 150 mm?
According to your curve now much force was the rubber band exerting when its length was 200 mm?
Now answer the three following questions, using the results you reported above:
What is the average of the forces exerted at 100 mm and 200 mm?
What is the average of the forces exerted at 100 mm and 150 mm?
What is the average of the forces exerted at 150 mm and 200 mm?
Place the answers to the six questions, one to a line, below. Starting in the 7th line give the units of your results and explain how you obtained them.
(note that your first answer should have been somewhere between 3 N and 10 N, your second between 3 N and 7 N, your third between 7 N and 10 N).
Would you expect the average of the forces exerted at 100 mm and 200 mm to be equal to the force exerted at 150 mm? Why or why not?
Your answer (start in the next line):
At the 100 mm point it’s about 3 N. At the 200 mm point its around 10 N. At the 150 mm point the force it’s about 7.5 N. The average force at 100 and 200 point is less than that of the 15o point. This graph looks like it is increasing at a decreasing rate. Is not linear.
Average force between 100 and 200= (3 +10)/2= 6.5 N The average of forces between 100 mm and 150 mm is around (3 N + 7. 5N) / 2 = 5.25. The average of the forces between 150 mm and 200 mm is around (7.5 N + 10 N) / 2 = 8.75 N
:
#$&* forces at 100,150,200 mm; ave 100 to 200, 100 to 150, 150 to 200 &&
Now answer the following:
Do you expect that the average force exerted between 100 mm and 200 mm is closer to the average of the two forces exerted at these points, or to the force exerted at 150 mm?
Do you expect that the average force exerted between 100 mm and 200 mm is greater or less than the average of the two forces exerted at these points?
Do you expect that the average force exerted between 100 mm and 200 mm is greater or less than the force exerted at 150 mm?
Give your answers, and the reasons for your answers, below:
Your answer (start in the next line):
The average of these forces would occur between 100 mm and 200 mm points on the graph (the midpoint between these two values.)
The graph is increasing at a decreasing rate and is not straight.
Its midpoint number is higher than the midpoint of the straight line.
#$&* 100 mm to 200 mm ave closer to ave or midpt force; 100 to 200 mm > or < ave of two; < or > force at 150 mm
Give on the first line below your best estimate for the average force exerted between the 100 mm and 200 mm lengths, and explain beginning on the second line how you obtained your estimate:
Your answer (start in the next line):
The average force is greater than the average of the 100 mm and 200 mm (should be somewhere in between the two.) You can tell because of the shape of the graph.
#$&* best estimate ave force 150 mm to 200 mm: &&
Now give in the first line your best estimate of the average force exerted between the 100 mm and 150 mm positions. In the second line do the same for the interval between the 150 mm and 200 mm positions.
Your answer (start in the next line):
I think that the average force will be more than the average of the 100 mm and 150 mm forces
#$&* best estimate ave. force 100 mm to 150 mm:, then 150 mm to 200 mm
Is the average of the two forces given in the preceding equal to, greater than, or less than the average given before that?
Should the average of the two forces given in the preceding be equal to, greater than, or less than the average force over the entire interval, or is it not possible to say?
Your answer (start in the next line):
Should be more than
#$&* ave of 100 mm to 150 mm force and 150 mm to 200 mm force compared to ave previously estimated for 100 mm to 200 mm:
The effect of a net force on the squared velocity of an object is equal to the average value of the force, multiplied by the distance through which it acts. (qualify... )
If the rubber band snaps back from the 150 mm to the 100 mm length while exerting its force on an object, the average force exerted on the interval from 100 mm to 150 mm lengths is exerted over a distance of 50 mm.
The analogous statement can be made if the rubber band snaps back from the 200 mm length to the 150 mm length.
What is the product of the average force and the distance for the first interval, from 100 mm to 150 mm lengths?
What is the product of the average force and the distance for the second interval, from 150 mm to 200 mm lengths?
Give your answers to these two questions in comma-delimited format in the first line below. In the second line give the units of your result and explain how the units were obtained. Starting on the third line explain how you got these answers.
There are two possible standard units for these calculations. The calculations you have likely made so far are not in either of these standard units; at least the units of the graph do not lead to standard units for force * distance.
However in what follows, until standard units are defined, use the units of each graph to do your calculations, and report all results in these units.
It will be easy to change these units to either of the standard units, and we will define the standard units and do some of these conversions near the end of this exercise.
You might or might not have included units in your preceding results.
What are the units of the average forces, according to the graph?
What are the units of the distances represented on the graph?
What therefore are the units of average force * distance as represented by areas on your graph?
Give one answer in each of the first three lines. Starting in the fourth line explain how you obtained your answer.
Your answer (start in the next line):
According to the graph, force is in Newtons, distance is in mm, product units of N * mm
#$&* units of average forces, distances, ave f * dist
According to your estimate of the average force between the 100 mm and 200 mm positions, what would be your force * distance result for this 100 mm interval?
How should your answer compare to the two answers given in the previous?
How nearly do your results actually compare and why, in terms of the way you arrived at your two results, would we in fact expect a modest discrepancy?
Give you answers and fully explain your reasoning below.
Your answer (start in the next line):
7.5 N * 100 mm = 750 N * mm
#$&* force * dist based on est ave force 100 mm to 200 mm, how should answer compare with two prev answers, how close and why is discrepancy expected
Based on the two intervals (100 mm to 150 mm and 150 mm to 200 mm) what is the total of your average force * displacement results?
Based on the single interval from 100 mm to 200 mm, what is the average force * displacement?
Give your two answers, in the order of the two questions, as two numbers in the first line below, in comma-delimited format.
Would you expect your answer based on the sum of the results for the two intervals to be more or less accurate than your result for the single interval from 100 mm to 200 mm? Why or why not?
Your answer (start in the next line):
7 N * 100 mm = 700 N * mm.
7.5 N * 100 mm = 750 N * mm.
I would think it would be more accurate having the single interval
#$&* total ave force * displacement 100 mm to 150 mm and 150 mm to 200 mm, previously given average based on single interval:, which should be more accurate:
Explain whether or not you think you could get a more accurate result by dividing the interval from 100 mm to 200 mm into four intervals (100 mm to 125 mm, 125 mm to 150 mm, 150 mm to 175 mm and 175 mm to 200 mm), and explain why you think as you do:
Your answer (start in the next line):
I think that the shorter the interval, the closer it would be.
Would have a more accurate answer from 150 to 175 (shorter)
if divided into four intervals how would accuracy be affected:
your brief discussion/description/explanation:
#$&*
According to your curve, at what length does the rubber band first begin to exert a force?
Give in the first line your estimate of average force * distance for the interval from this length to the 200 mm length. Do your best to make your estimate accurate to within 3%.
Explain in the second line whether you think it is feasible, with care, to answer within 3%. Explain also how you would proceed if you wanted to make the estimate accurate to within 1%.
Explain how you obtained your force * distance result and why you think it is within 3%.
Your answer (start in the next line):
It looks like it approaches zero around 80 mm
#$&* ave force 3%, why 3% how to get 1%, how 3%
The figure below doesn't represent a rubber band, which exerts a variable force, not a constant force.
In the figure below the force is in Newtons and the position is in centimeters. The force is exerted from a position of x = 8.0 cm to x = 9.0 cm.
The figure below represents another force, double the first, exerted between positions x = 8.0 cm and x = 9.0 cm.
The figure below represents the original force, exerted between positions x = 8.0 cm and x = 10.0 cm.
Answer the following:
How does the average force * distance for the interval covered by the second graph compare to that of the first?
How does the average force * distance for the interval covered by the third graph compare to that of the first?
How does the average force * distance for the interval covered by the third graph compare to that of the second?
Explain in detail how you reasoned out your conclusions.
Your answer (start in the next line):
first graph: the force is between 1 N and 1.5 N. second graph: force is between 2.5 N and 3 N. third graph: force is between 1 N and 1.5 N, third graph is longer than the others
Looks like the force * distance by the second graph is about double the first. The third graph distance is longer and is about 2 cm, so the force * distance represented by the third graph is about double the first.
third graph is about the same as the first
#$&* ave F ds 2d to 1st, 3d to 1st, 3d to 2d
Now answer the following:
How does the average force represented in the graph below compare to that of the first graph shown in the series above?
How does the distance represented in the graph below compare to that of the first graph shown in the series above?
How does the average force * distance for the interval depicted below compare to that of the first force shown above?
Answer these questions below, and explain how this series of graphs and questions has clarified the statement that 'twice the force exerted through twice the distance has four times the effect'.
Your answer (start in the next line):
The average force is about double the first graph.
this graph to the second graph given earlier, double the force * distance of first. force * distance on this graph is double that of the second graph, The force * distance is therefore 4 times as great as the first.
#$&* new graph to 1st graph ave force, dist, ave force * dist
The 'effect' we've been talking about above is the effect of a net force on the squared velocity of a given object.
Specifically we have the following:
If we multiply the average net force exerted on an object of mass m by the displacement of that object in the direction of the force, the result we obtain is called the 'net work' done on the object. The effect of this net work will be to increase the quantity 1/2 m v^2 by an amount equal to the net work. 1/2 m v^2 is called the kinetic energy of the object, and the statement that 'net work is equal to change in kinetic energy' is called the work-energy theorem.
This is a fairly simple statement, but it has wide-ranging implications and it takes a lot of work with a wide variety of examples to fully appreciate and understand its meaning and implications.
Within the context of rubber bands, we use the following terminology:
The product of the average tension exerted by a rubber band as it is stretched from one length to another, multiplied by the distance through it is stretched, is the work-energy associated with that stretch.
The term 'work-energy' indicates that 'work' and 'energy' are equivalent and pretty much interchangeable terms (but to avoid possible confusion we do have to learn to be careful how we use those terms).
Most of the energy associated with the stretching of the rubber band can be recovered when the rubber band is allowed to 'snap back'. However as a rubber band stretches or snaps back, there are complex effects involving thermal energy (heating and cooling) and a significant amount of the energy ends up being converted to thermal energy and dissipated.
An 'ideal rubber band' is one in which thermal energy losses are negligible. There is no such thing as an ideal rubber band, and while real rubber bands are not all that far from the ideal, they aren't all that close either. The more common terminology used in physics is that of an 'ideal spring'; this is because metal springs are more often used in experiments and applications and typically have much smaller thermal losses than rubber bands. Ideal springs also have linear force vs. length graphs, and though no actual spring results in perfect linearity, metal springs typically come much closer to this ideal than rubber bands.
Rubber bands are used here for several reasons:
They are cheaper.
They are typically lighter and interfere with certain other physical properties of a system (e.g., mass and weight) less than do metal springs.
They clearly exhibit non-ideal behavior at a directly observable level.
All this leads up to a couple of very important statements:
When a rubber band is stretched the energy associated with the process is equal to the average force of tension multiplied by the distance of the stretch.
This quantity is equal to the area beneath a graph of force vs. length.
Most of this energy can be recovered when the rubber band snaps back.
The recoverable energy is called the potential energy of the rubber band.
So increasing the length of a rubber band increases its potential energy by an amount which is approximately equal to (but a bit less than) the corresponding area beneath its tension vs. length graph.
Using the new terminology, answer the following, assuming that the rubber band of the original graph does not lose any energy to thermal effects:
By how much does its potential energy increase as it is stretched from a length of 100 mm to a length of 150 mm?
By how much does its potential energy increase as it is stretched from a length of 150 mm to a length of 200 mm?
By how much does its potential energy increase as it is stretched from a length of 100 mm to a length of 200 mm?
How much energy would we expect to get back if the rubber band 'snapped back' from its 200 mm length to a length of 150 mm?
How would this last result be obtained from your answers to the first and third questions in this series?
Answer with four numbers in comma-delimited format in the first line, followed by an explanation starting at the second line.
Your answer (start in the next line):
Between 100 mm and 150 mm the average force * distance is probably between 5.5 N * 50 mm and 6 N * 50 mm, putting it between 275 N * mm and 300 N * mm.
Between 150 mm and 200 mm the average force * distance is about 8 N * 50 mm and 9 N * 50 mm, 400 N * mm and 450 N * mm.
potential energy change for the interval from 100 mm to 200 mm is the sum of the potential energy change from 100 mm to 150 mm
work done = work required to stretch it
#$&* `dPE 100 to 150 mm, 150 to 200 mm, 100 to 200 mm, energy back 200 to 150 mm, last related to 1st and 3d
How would the answers for a real rubber band, with its non-ideal thermal behavior, differ from those you gave? Would any of your answers not differ?
Your answer (start in the next line):
When a 'real' rubber band snaps back, some of the energy goes into heating and not returned
#$&* compare previous to behavior of real rb
**&@ If you haven't been assigned the rubber band calibration, construct a graph for the following data and use it for these questions:
rubber band first exerts measurable tension at length 8.1 cm
when supporting two 20-gram dominoes length is 8.4 cm
when supporting four 20-gram dominoes length is 8.8 cm
when supporting six 20-gram dominoes length is 9.1 cm
when supporting eight 20-gram dominoes length is 9.3 cm
when supporting ten 20-gram dominoes length is 9.5 cm
Using your data from the rubber band calibration experiment, answer the following questions for your first rubber band:
At what length, in cm, did it begin to exert a noticeable tension force? Place this answer in the first line below.
Between this length and .5 cm greater than this length, what was the average force in Newtons, and how much work was done between these two lengths (i.e., what is the corresponding area beneath the curve of your graph)? Place these two numbers in your second line, separated by commas.
Starting in the third line indicate the units of the area beneath the curve, and explain how you determined this area. Include the relevant data from your rubber band calibration experiment and explain how it was used:
Your answer (start in the next line):
rubber band was stretched .5 cm has .6 N
average force (0 N + .6 N) / 2 = .3 N, force * distance total of .3 N * .5 cm = .15 N * cm.
Final estimate .17 N * cm.
Graph increasing at decreasing rate
#$&* for your rb length measurable tension, ave F and `dW next .5 cm
Still using your information for the first rubber band:
How long was the rubber band when it supported the weight of two dominoes?
Between the length at which the rubber band started exerting force, and the length at which it supported the weight of two dominoes, what is the area beneath your curve?
Between the length at which the rubber band supported the weight of two dominoes, and the length at which it supported the weight of four dominoes, what is the area beneath your curve?
Between the length at which the rubber band supported the weight of four dominoes, and the length at which it supported the weight of six dominoes, what is the area beneath your curve?
Between the length at which the rubber band supported the weight of six dominoes, and the length at which it supported the weight of eight dominoes, what is the area beneath your curve?
Between the length at which the rubber band supported the weight of eight dominoes, and the length at which it supported the weight of ten dominoes, what is the area beneath your curve?
Give your answers as six numbers in the first line below, separated by commas. Starting in the 7th line explain how you obtained your results.
Your answer (start in the next line):
rubber band was 8.3 cm long
6 dominoes has around 1.3 N force and 8.6 cm long with 8 dominoes (1.6 N).
(1.3 N + 1.6 N) / 2 = 1.45 N
The distance would be 8.6 cm - 8.3 cm = .3 cm
area = average force * distance
#$&* lgth supporting 2 dom, area of region init to 2 dom, 2 to 4 dom, 4 to 6, 6 to 8, 8 to 10
The five areas you obtained above are the energies associated with stretching the rubber band between the various lengths. Answer the following:
What is the total energy stored in the rubber band, as a result of stretching it from the length at which it began to exert a tension force and the length at which it supported the weight of ten dominoes? This is the total energy of the rubber band at this length.
What would be the total energy of the rubber band at the length at which is supported six dominoes?
Be sure to explain the rationale for your answers.
Your answer (start in the next line):
Would need to stretch through each of the five intervals to get to 10 cm
70 N
75 N
80 N
95 N
120 N
#$&* total energy 10 dom, total energy 6 dom
Suppose you suspend six dominoes from the rubber band, which stretches the rubber band to the corresponding length (call this the 'first length'). Then you pull down on the dominoes until the rubber band has stretched to the same length it had when it supported ten dominoes (call this the 'second length').
Assuming no energy losses due to friction or to thermal effects:
How much work would you have to do on the rubber band to accomplish this stretch?
How much more energy does the rubber band have at the second length than at the first?
If you released the dominoes, how much energy would the rubber band lose between release and the instant it reaches the first length?
How much kinetic energy would the dominoes gain in between these instants?
What would therefore be the kinetic energy of the dominoes at the instant the rubber band reaches its first length?
Answer with five numbers in comma-delimited format in the first line, followed by an explanation in the second line.
Your answer (start in the next line):
The amount of work is the same as the difference in these energies. Energy is released when the rubber band snaps back (kinetic energy.) Would be an increase in kinetic energy and decrease The decrease in potential energy.
#$&* work 6 to 10 dom, how much more energy 10 than 6 dom, energy lost 10 to 6 dom, energy attained at 6 dom rel from 10, KE at 6 dom
F = m a, so units of force are equal to units of mass, multiplied by units of acceleration.
A Newton is a kg * m / s^2, and a dyne is a gram * cm / s^2.
How many grams are in a kg, and by what number would you multiply a quantity in kg m/s^2 to express the same quantity in gram * m / s^2? Give your comma-delimited answers in the first line.
How many cm are in a meter, and by what number would you multiply a quantity in gram m/s^2 to express the same quantity in gram * cm / s^2? Give your comma-delimited answers in the second line.
By what number would you therefore multiply a quantity in kg m/s^2 to express the same quantity in gram cm / s^2? Answer in the third line.
By what number would you multiply a quantity in gram cm / s^2 to get the same quantity in kg m/s^2? Answer in the fourth line.
Starting in the fifth line explain your reasoning.
Your answer (start in the next line):
kg is 1000 grams, kg m/s^2 is 1000 g m/s^2.
Meter =100 cm, g m/s^2 is 100 g cm /s^2.
1000 g m/s^2 is 100,000 g cm/s^2, or 10^5 g cm/s^2.
A Joule is a N * m, and an erg is a dyne * cm.
#$&* g in kg & mult kg m/s^2 to g m/s^2, cm in m & mult g m/s^2 to g cm/s^2, mult kg Your instructor is trying to gauge the typical time spent by students on these experiments. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:
Approximately how long did it take you to complete this experiment?
Your answer (start in the next line):
3 hours"
&#This looks good. Let me know if you have any questions. &#