#$&*
phy 201
Your 'energy conversion 1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Energy Conversion 1_labelMessages **
7/10 5pm
** **
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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):
:force at length 100 mm:3N
:force at length 150 mm:7N
:force at length 200 mm:10N
average of forces 100 mm to 200 mm:6.5N
average of forces 100 mm to 150 mm:5.5N
average of forces 150 mm to 200 mm:8.5N
explanation: adding my estimates up and dividing by 2.
expect ave of 100 mm and 200 mm forces equal to 150 mm force? No
#$&* forces at 100,150,200 mm; ave 100 to 200, 100 to 150, 150 to 200 &&
Now answer the following:
Do you expect that the average force exerted between 100 mm and 200 mm is closer to the average of the two forces exerted at these points, or to the force exerted at 150 mm?
Do you expect that the average force exerted between 100 mm and 200 mm is greater or less than the average of the two forces exerted at these points?
Do you expect that the average force exerted between 100 mm and 200 mm is greater or less than the force exerted at 150 mm?
Give your answers, and the reasons for your answers, below:
Your answer (start in the next line):
average force between 100 mm and 200 mm closer to the average of the two force, or to the force exerted at 150 mm: closer to the force at 150mm
average force between 100 mm and 200 mm greater or less than ave of the two forces: more than the average
average force between 100 mm and 200 mm greater or less than force exerted at 150 mm: less than 150mm
explanation: With The forces increments from 3-7-10, since this is increasing at a decreasing rate, I believe it will be less than the 150mm force, but over the average.
#$&* 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):
6.8N
The average force of 100mm and 200mm is 6.5N. The force at 150mm is 7N. From my prediction that the force would be under but closer to the 150mm force, I took a guess.
#$&* best estimate ave force 150 mm to 200 mm: &&
Now give in the first line your best estimate of the average force exerted between the 100 mm and 150 mm positions. In the second line do the same for the interval between the 150 mm and 200 mm positions.
Your answer (start in the next line):
best estimate of ave force 100 mm to 150 mm: 5.65N
best estimate of ave force 150 mm to 200 mm: 8.6N
#$&* 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):
Is the average of the two forces greater or less than the best-estimate average you gave previously for the force between 100 mm and 200 mm:greater than
Should these two averages be equal?i think that the actual average would be slightly more than the average I have calculated.
#$&* ave of 100 mm to 150 mm force and 150 mm to 200 mm force compared to ave previously estimated for 100 mm to 200 mm:
The effect of a net force on the squared velocity of an object is equal to the average value of the force, multiplied by the distance through which it acts. (qualify... )
If the rubber band snaps back from the 150 mm to the 100 mm length while exerting its force on an object, the average force exerted on the interval from 100 mm to 150 mm lengths is exerted over a distance of 50 mm.
The analogous statement can be made if the rubber band snaps back from the 200 mm length to the 150 mm length.
What is the product of the average force and the distance for the first interval, from 100 mm to 150 mm lengths?
What is the product of the average force and the distance for the second interval, from 150 mm to 200 mm lengths?
Give your answers to these two questions in comma-delimited format in the first line below. In the second line give the units of your result and explain how the units were obtained. Starting on the third line explain how you got these answers.
Your answer (start in the next line):
your brief discussion/description/explanation: ds*FAve=net force
product of ave force and distance, interval 100 mm to 150 mm:50mm*5.5N=275N
@&
50 mm * 5.5 N = 275 N * mm, not 275 N.
*@
product of ave force and distance, interval 150 mm to 200 mm:50mm*8.5N=425N
#$&* ave force * dist 100 mm to 150 mm, ave force * dist 150 mm to 200 mm:
Note the following about units:
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):
units of forces:N
units of distances:mm
units of ave force * distance:N*mm
#$&* units of average forces, distances, ave f * dist
According to your estimate of the average force between the 100 mm and 200 mm positions, what would be your force * distance result for this 100 mm interval?
How should your answer compare to the two answers given in the previous?
How nearly do your results actually compare and why, in terms of the way you arrived at your two results, would we in fact expect a modest discrepancy?
Give you answers and fully explain your reasoning below.
Your answer (start in the next line):
force * distance for 100 mm interval:680N*mm
how answer should compare to two previous: this would be the sum of the forces
how close in actuality:if we add the previous forces up we get 700N*mm, this is 20N*mm more than the value just calculated.
why we would expect a discrepancy: the graph is not linear, had the graph been linear we could take the average force 100mm and 200mm to find the force at 150mm. instead we are playing best guess.
#$&* 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):
total of ave force * displacement based on 2 intervals:700N*mm
total ave force * displacement based in single interval:680N*mm
#$&* total ave force * displacement 100 mm to 150 mm and 150 mm to 200 mm, previously given average based on single interval:, which should be more accurate: I think the 2 intervals added together would be more accurate.
Explain whether or not you think you could get a more accurate result by dividing the interval from 100 mm to 200 mm into four intervals (100 mm to 125 mm, 125 mm to 150 mm, 150 mm to 175 mm and 175 mm to 200 mm), and explain why you think as you do:
Your answer (start in the next line):
if divided into four intervals how would accuracy be affected: the accuracy would be greater
your brief discussion/description/explanation: with each interval the force changes, the more intervals we can calculate the force of, the better we can measure them.
#$&*
How division into four intervals would affect accuracy:
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):
Length at which rubber band began to exert force:80mm
Ave force * distance to 200 mm length within 3%:600N*mm
Explanations:(F0+Ff)/2*120mm=5N*120mm=600N*mm. We could make this more accurate if we were using more than the one interval.
#$&* 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.
Id 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.1.3N*cm

The figure below represents another force, double the first, exerted between positions x = 8.0 cm and x = 9.0 cm. =2.6N*cm

The figure below represents the original force, exerted between positions x = 8.0 cm and x = 10.0 cm.2.6N*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):
Comparison second graph to first:1st:1.3*1=1.3N*cm 2nd:2.6*1cm=2.6N*cm. net force is double on the second due to double the FAve
Comparison third graph to first:3rd: 1.3*2=2.6N*cm net force is double on the second due to double the distance
Comparison third graph to second: net force is the same, although the FAve and distance change between both, they both multiply to the same amount.
Explanations:
#$&* 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):
Comparison of new graph to first in preceding series: comparison of ave forces: 2.6N/1.3N=2..2*
Comparison of new graph to first in preceding series: comparison of distances: 2cm/1cm=2..2*
Comparison of new graph to first in preceding series: comparison of ave force * distance: 5.2Ncm/1.3Ncm=4.. 4*
#$&* 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):
PE increase of rubber band 100 mm to 150 mm length: 275N*mm
PE increase of rubber band 150 mm to 200 mm length: 425N*mm
PE increase of rubber band 100 mm to 200 mm length: 700N*mm
Energy expected snapping back from 200 mm to 150 mm length: 425N*mm
How last answer would be determined from first and third answers:700N*mm-275N*mm
Explanations: In an ideal system, the rubber band would snap-back with the same amount of energy taken to get it to this point. Some of this energy is lost in heat.
#$&* `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):
The PE should stay the same, but the KE will have some losses in heat.
Which answers would differ and which would not for real as opposed to ideal rubber band:
#$&* compare previous to behavior of real rb
**&@ If you haven't been assigned the rubber band calibration, construct a graph for the following data and use it for these questions:
rubber band first exerts measurable tension at length 8.1 cm
when supporting two 20-gram dominoes length is 8.4 cm
when supporting four 20-gram dominoes length is 8.8 cm
when supporting six 20-gram dominoes length is 9.1 cm
when supporting eight 20-gram dominoes length is 9.3 cm
when supporting ten 20-gram dominoes length is 9.5 cm
Using your data from the rubber band calibration experiment, answer the following questions for your first rubber band:
At what length, in cm, did it begin to exert a noticeable tension force? Place this answer in the first line below.
Between this length and .5 cm greater than this length, what was the average force in Newtons, and how much work was done between these two lengths (i.e., what is the corresponding area beneath the curve of your graph)? Place these two numbers in your second line, separated by commas.
Starting in the third line indicate the units of the area beneath the curve, and explain how you determined this area. Include the relevant data from your rubber band calibration experiment and explain how it was used:
Your answer (start in the next line):
11.3cm
.475N, .2375N*cm
(.2N+.75N)/2=.475N
475N(.5cm)=.2375N*cm
length at which tension force became noticeable, real rubber band:
ave force in Newtons between this length and .5 cm greater length:
units of area beneath curve, explanation:
#$&* 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?
Your answer (start in the next line):
2-domino length:11.5cm
area up to 2-domino length:.06N*cm
area between 2- and 4- domino lengths:.55N*cm
area between 4- and 6- domino lengths:.95N*cm
area between 6- and 8- domino lengths:1.33N*cm
area between 8- and 10- domino lengths:1.7N*cm
explanations:dF/dL=netF
#$&* lgth supporting 2 dom, area of region init to 2 dom, 2 to 4 dom, 4 to 6, 6 to 8, 8 to 10
The five areas you obtained above are the energies associated with stretching the rubber band between the various lengths. Answer the following:
What is the total energy stored in the rubber band, as a result of stretching it from the length at which it began to exert a tension force and the length at which it supported the weight of ten dominoes? This is the total energy of the rubber band at this length.
What would be the total energy of the rubber band at the length at which is supported six dominoes?
Be sure to explain the rationale for your answers.
Your answer (start in the next line):
total energy at max length:4.59N*cm
total energy at 6-domino length:1.56
explanations:The sum of all energies to get the 10-domino energy. Sum of energies up to the 4-6 dominos.
#$&* 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?
Your answer (start in the next line):
work to extend rb supporting 6 dominoes to length at which it would support 10 dominoes:3.03N*cm
how much more energy at second length than at first:3.03N*cm
if dominoes released, energy rubber band would lose in contracting to first length:3.03N*cm
KE gained by dominoes:3.03N*cm
KE of dominoes when first length reached:0N*cm
#$&* work 6 to 10 dom, how much more energy 10 than 6 dom, energy lost 10 to 6 dom, energy attained at 6 dom rel from 10, KE at 6 dom
Your areas, and hence the units in which you have reported work and energy to this point, should be Newton * mm (the units of the first graph are Newtons of force vs. millimeters of length) and Newton * cm (the units of your calibration graphs should be Newtons of force vs. cm of length).
Newton * mm and Newton * cm are both valid units of energy, but neither is a standard unit.
There are two standard units for energy. One is the Newton * meter, also called a Joule, and you are likely familiar with this unit. The other is the dyne * cm, also called an erg.
How many mm are in a meter? How many Newton * mm are therefore in a Newton * meter or Joule? Give your comma-delimited answers in the first line.
By what number would you therefore multiply a quantity given in N * m in order to get the same quantity in N * mm? And by what number would you multiply a quantity in N * mm to get the same quantity in N * m or Joules? Give your comma-delimited answers in the second line.
How many cm are in a meter? How many Newton * cm are therefore in a Newton * meter or Joule? Give your comma-delimited answers in the third line.
By what number would you therefore multiply a quantity given in N * m in order to get the same quantity in N * cm? And by what number would you multiply a quantity in N * cm to get the same quantity in N * m or Joules? Give your comma-delimited answers in the fourth line.
Starting in the fifth line give an outline of the reasoning used to compare these units.
Your answer (start in the next line):
1000, 1000
1000, .001
100, 100
100, .01
these are standard units of conversion.
#$&* mm in m & N mm in N m, mult N m to N mm ^ N mm to N m, cm in m & N cm in N m, multipliers
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):
1000, 1000
100, 100
100000
.00001
#$&* g in kg & mult kg m/s^2 to g m/s^2, cm in m & mult g m/s^2 to g cm/s^2, mult kg m/s^2 to g cm/s^2, vice versa
From the above you should see that a dyne is 1/100,000 N, or 10^-5 N, while a Newton is 100,000 dynes, or 10^5 dynes.
Recall that a Joule is a Newton * meter, and an erg is a dyne * centimeter.
How many ergs does it therefore take to make a Joule?
By what number would you multiply work in Joules to get work in ergs?
By what number would you multiply work in ergs to get work in Joules?
Give your comma-delimited answers in the first line.
Explain your thinking starting in the second line.
Your answer (start in the next line):
10000000 , .0000001 , 10000000
#$&* ergs in J, mult J to erg, mult erg to J
How much energy, in Joules, does your first rubber band store when it is at the length required to support 10 dominoes?
How many ergs is this?
Give your two answers in comma-delimited format in the first line below, and your explanation starting in the second line:
Your answer (start in the next line):
.0459 J, 459000
#$&*
energy in J at 10 dom lgth, same in ergs
Your instructor is trying to gauge the typical time spent by students on these experiments. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:
Approximately how long did it take you to complete this experiment?
Your answer (start in the next line):
2.5hrs
#$&*
self-critique rating
&#Very good data and responses. Let me know if you have questions. &#