Calculating the elastic potential energy from a force-extension investigation

[lpettigrew]

Homework Statement

Hello, I have a question that I am struggling to answer. I have attempted all of the sections but would greatly appreciate any help or guidance in areas where I am clearly confused.

A student demonstrates an experiment to analyse the force-extension characteristics of a rubber band.
1.Describe the equipment the student could use to perform the experiment.
2. Below are the students measurements, Use this information to plot a graph of force versus extension.
3. The student wants to calculate the elastic potential energy stored by using the equation EPE = (1/2)kx^2. Why would she get the wrong answer? Would it be an underestimate or an overestimate?
4. Describe the energy transfer as the force is increased, then decreased. How is this represented in your graph?

Relevant Equations

Eel=1/2FΔx

1. The student should use a rubber band, g-clamp, a retort stand, boss and clamp, a mass hanger, 100g masses and a metre rule.

The rubber band should be positioned to hang freely from the retort stand, held in place by a g-clamp to the laboratory bench. Measure the length of the rubber band using a ruler. At the base of the rubber band a mass holder should be looped across to hang below and a metre rule should be used to measure the new length of the band. The student should gradually increase the load on the band, recording the change in the length of the rubber band.
The student may wish to repeat the investigation with a selection of rubber bands.

2. See attached

3. This is where the greatest confusion lies. Since, I know that Eel = 1⁄2FΔx. Since Hooke’s law states ΔF = kΔx, substituting the expression for ΔF into the equation for the elastic strain energy equation produces the formula ΔEel = 1⁄2k(Δx)^2.
On the force extension graph, the area between the line on the force-extension graph and the extension axis will represent the work done, and thus the elastic strain energy. This is only true for linear relationships though, like Hooke's Law, as the area under the line is a triangle and the formula for the area of a triangle, 1/2 bh, is equivalent to the formula for the elastic strain energy, Eel=1/2Fx
Therefore, as a rubber bands follows a hysteresis loop and has a non-linear relationship, the work done (and hence the Eel) can be found from the area under the line. However, you cannot use the formula 1/2Fx to calculate this as this is the area of a triangle, where the area underneath the hysteresis loop of the rubber band is not a triangle. Usually, if a non-linear force-extension graph has a curved line, finding the area may require one to estimate or count the squares of graph paper under the line. In such an instance, one is further required to multiply the number of squares by the elastic strain energy value (Fx) for each square. The student wishes to calculate the elastic potential energy stored, which is equal to the area under the loading curve of the force-extension graph. I am not sure whether this will be an under or overestimate though?

4. I am not too sure how to answer here. Would it be that energyis absorbed and stored when a the force is increased and the rubber band is stretched but the energy is released when the force is decreased. Rubber absorbs more energy during loading than it releases in unloading. This difference is represented by the area of the hysteresis loop, between the curves of loading and unloading.
The effect of hysteresis in rubber is to transfer energy to its molecules, resulting in heating.

Could I elaborate or improve upon my answer to be more concise ?

I greatly appreciate any help

 

[BvU]

2. See attached

Doesn't work for me ?!

 

[kuruman]

Doesn't work for me ?!

Ditto.

 

[haruspex]

not sure whether this will be an under or overestimate though?

Your analysis for part 3 is good, but I am puzzled by your question at the end. You seem to be asking whether counting squares will give an underestimate or overestimate. The question posed is whether $\frac 12kx^2$ will give an underestimate or overestimate... but it does not specify how that k is to be determined. Basing k on the slope at the high end will give a different result from basing it on the whole extension. I suggest you pick a method, state it, and base your answer on that.

Re 4, not sure how to say any more without data on how some energy goes into heating. Does it happen mostly during stretch, during release, or roughly equally? Seeing the actual curve might give a hint.

 

[lpettigrew]

Sorry I forgot to add it, sorry the graph is quite crude.
@kuruman and @BvU @haruspex

 

[lpettigrew]

Your analysis for part 3 is good, but I am puzzled by your question at the end. You seem to be asking whether counting squares will give an underestimate or overestimate. The question posed is whether $\frac 12kx^2$ will give an underestimate or overestimate... but it does not specify how that k is to be determined. Basing k on the slope at the high end will give a different result from basing it on the whole extension. I suggest you pick a method, state it, and base your answer on that.

Re 4, not sure how to say any more without data on how some energy goes into heating. Does it happen mostly during stretch, during release, or roughly equally? Seeing the actual curve might give a hint.

Thank you for your reply, for question 3 I meant that I was not sure how firstly whether counting the number of squares would be an appropriate method to find the area under the curve?
Secondly, I do not know whether the student would get an under or overestimate using the forumla EPE=1/2kx^2, nor am I sure how to test this. Would I as you suggest find the gradient at a point on the curve to find the spring constant and use this to find the area using 1/2 kx^2. Then, find the actual area underneath the loading curve and find the difference between the two?
For example, you say find k at the high end of the loading curve. The gradient between the final two points on the loading graph is actaully a straight line, so could I just use Δy/Δx to find k. Δy/Δx=12-10/15-11.8=2/3.2=0.625

EPE=1/2 *0.625 * 15 cm ^2=70.3 J

Whereas, then should I evauate the graph to find the area underneathed the loading curve to find the EPE by counting the number of squares and mutliplying these by the elastic strain value (F*Δx) for each square?

Sorry, I really I am quite confused.

For question 4 I have attached the student's results and my graph to help visualise this.

Thank you again

 

[haruspex]

whether counting the number of squares would be an appropriate method to find the area under the curve?

Yes, that's a fairly standard way of getting areas from graphs. Where the line goes through a square you 'round to nearest'.

I do not know whether the student would get an under or overestimate using the forumla EPE=1/2kx^2, nor am I sure how to test this. Would I as you suggest find the gradient at a point on the curve to find the spring constant and use this to find the area using 1/2 kx^2. Then, find the actual area underneath the loading curve and find the difference between the two?
For example, you say find k at the high end of the loading curve. The gradient between the final two points on the loading graph is actaully a straight line, so could I just use Δy/Δx to find k. Δy/Δx=12-10/15-11.8=2/3.2=0.625

First, there is an ambiguity in the question regarding "stored potential energy". Are we to take that as the work put in or the work that can be got back out?
For finding a value to use for k, yes it could be done as you say but I now think a more reasonable way is to take it as the best fit slope to all the datapoints. So that should put a line roughly through the middle of the hysteresis. Whether that produces an underestimate or overestimate depends on how stored PE is defined.

Re "how represented" in 4, still not sure what they want. Maybe just say work transferred in is area under upper line, work transferred out is area under lower line.

 

[kuruman]

A creative way to find the area under the curve, if you have a reasonably sensitive weighing scale such as a digital kitchen scale, is to
1. Make a photocopy of your graph.
2. Cut a rectangular piece (away from the lines) and find its area in Joules considering the scaling of your axes.
3. Cut the area under the loading curve, weigh and convert to Joules.
4. Cut the area under the unloading curve, weigh and convert to Joules.

This method assumes that the paper has uniform density and thickness which is reasonable. If you are concerned about accuracy and sensitivity, use more than one photocopies and stack them. This method beats counting squares, especially if they are small, not to mention trying to figure out the fraction of a square that is under the curve.

 

[lpettigrew]

Yes, that's a fairly standard way of getting areas from graphs. Where the line goes through a square you 'round to nearest'.

First, there is an ambiguity in the question regarding "stored potential energy". Are we to take that as the work put in or the work that can be got back out?
For finding a value to use for k, yes it could be done as you say but I now think a more reasonable way is to take it as the best fit slope to all the datapoints. So that should put a line roughly through the middle of the hysteresis. Whether that produces an underestimate or overestimate depends on how stored PE is defined.

Re "how represented" in 4, still not sure what they want. Maybe just say work transferred in is area under upper line, work transferred out is area under lower line.

@haruspex Thank you again for your reply. Following your advice, I have added a line of best fit to my curve, although I do no know whether this is correct (i have attached another picture here) . I have tried to distribute the data points evenly on either side and position it to run roughly through the middle of the hysterisis as you state.
The gradient of this line would be equal to;
By selecting two random points (9.5,6) and (4.2,0)
Δy/Δx=6-0/9.5-4.2=6/5.3~1.13
Therefore, k=1.13
Would I be able to use the formula, EPE=1/2kx^2, to find the EPE, as I have stated that this can only be applied to linear relationships where the force-extension graph is a straight line, not a curve as in the case of the hysteresis loop of the rubber band. Or since I have drawn a line of best fit, which follows a linear relationship, can this equation be used?
Also, what value of x do I input, would this be the maximum extension of 15cm?
In which case; EPE=1/2*1.13*15^2=127.125 J
This would be an underestimate because the line of best fit falls below the curve of the graph when the band is loaded to exhibit the stored EPE. Therefore, the area below the loading curve will actaully be greater, and thus a greater EPE.

Question 4 Thank you yes I think that I will again take your advice here. Perhaps it is also worth mentioning that this forms a hysterisis loop and maybe it is worth my stating what that is ?

 

[lpettigrew]

A creative way to find the area under the curve, if you have a reasonably sensitive weighing scale such as a digital kitchen scale, is to
1. Make a photocopy of your graph.
2. Cut a rectangular piece (away from the lines) and find its area in Joules considering the scaling of your axes.
3. Cut the area under the loading curve, weigh and convert to Joules.
4. Cut the area under the unloading curve, weigh and convert to Joules.

This method assumes that the paper has uniform density and thickness which is reasonable. If you are concerned about accuracy and sensitivity, use more than one photocopies and stack them. This method beats counting squares, especially if they are small, not to mention trying to figure out the fraction of a square that is under the curve.

@kuruman Thank you for your reply. That is such an innovative and fantastic idea! I desperately want to try it now, I photocopied my graph and produced a few spares to stack as suggested. However, I found the battery in my kitchen scales is dead, so I will have to wait until tomorrow to get a new one and try this. Honestly, I will try it with other graphs also since I am captivated by the idea so thank you very much I am extremely grateful

 

[haruspex]

what value of x do I input, would this be the maximum extension of 15cm?

Yes.

 

[kuruman]

@kuruman Thank you for your reply. That is such an innovative and fantastic idea! I desperately want to try it now, I photocopied my graph and produced a few spares to stack as suggested. However, I found the battery in my kitchen scales is dead, so I will have to wait until tomorrow to get a new one and try this. Honestly, I will try it with other graphs also since I am captivated by the idea so thank you very much I am extremely grateful

Please let us know how it turned out. I would be interested to know how the weighing method compares with the squares counting method.

 

[lpettigrew]

Yes.

@haruspex Thank you for your reply, so is the rest of my answer for question 3 correct too?

 

[lpettigrew]

Please let us know how it turned out. I would be interested to know how the weighing method compares with the squares counting method.

Yes, I will get a battery today and let you know how I get on

 

[haruspex]

@haruspex Thank you for your reply, so is the rest of my answer for question 3 correct too?

As I wrote, it is unclear what is meant by the potential energy stored. If it means what was put in the method gives an underestimate, but for what can be got back out it is an overestimate.

 

[lpettigrew]

As I wrote, it is unclear what is meant by the potential energy stored. If it means what was put in the method gives an underestimate, but for what can be got back out it is an overestimate.

Oh of course, sorry I had neglected that thought. That makes absolute sense. Thank you very much again for all of your help.

 

[kuruman]

Isn't the fact that the work done by the force over a closed loop is not zero ($\oint\vec F\cdot d\vec s\neq 0$) sufficient to conclude the $\vec F$ is non-conservative and, therefore, not derivable from a potential energy? Of course, one may model the elastic force as the sum of a conservative term that obeys Hooke's law and a non-conservative term that will more or less produce the hysteresis effect. Then the "potential energy stored" would be easily calculable.

I think that, for any model to make sense and be useful, at zero load it should predict the unstrained length of the elastic band. For this reason, the straight line in post #9 does not look right to me.

Another observation I have is about procedure. Did you wait some time (~10 min) before measuring the distance after changing a particular load? Elastic bands do not reach their final length instantly.