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13 August 2014

AP Physics 1: Define the system carefully when doing energy problems

For years I've taught energy conservation via an equation.  Write the energy conservation equation for the object in question, identify each term, and solve.  That gets the right answer most of the time.  However, in AP Physics 1, getting the right answer is not as important as describing clearly how to get the right answer, and why the right answer is in fact the right answer.  I need to dig deeper for this new course.

In particular, I've always been fluffy about defining the system experiencing these energy changes.  I implicitly was referring to the single object experiencing a change in position or velocity.  That's "fluffy" because potential energy is created through an interaction within a system of multiple objects; and because I'm making all sorts of assumptions about what forces doing work should be included on the left side of this equation.  The students could use this equation to get the right answer, but they could not have described in careful, correct language the energy transfer.

The NTIPERS book provided me with a guide to teaching my students to use a deeper approach.  In each energy problem presented by this book, the authors clearly define the system being analyzed, and they list the forms of energy the system could possibly possess before, during, and after the process in question.  NTIPERS suggests making a bar graph for each problem listing the amount of each energy; words and numbers are probably enough, but the bar graph is a nice approach, too.

As an example: consider 50 kg Tarzan dropping on a vine from rest, as on the 2014 AP Physics B exam problem 1.  One part asks the student to calculate the speed of Tarzan as he swings through his lowest position (call it 2 m below his starting point), neglecting air resistance.  We can approach this two equivalent ways:

1. Treating the Tarzan-Earth system:  Systems can have potential energy due to their interaction.  Systems can also have kinetic energy, which is equal to the sum of the kinetic energies of the component parts of the system.  Define the lowest position of Tarzan as the zero of gravitational energy for the system.  That means the Tarzan-Earth system begins with mgh = 1000 J of gravitational energy, and (because nothing is moving) no kinetic energy, giving 1000 J of system mechanical energy at the start.  During the process, no work is done by an object external to the Tarzan-Earth system -- the only other object exerting a force on any part of the system is the rope, and the rope is always pulling perpendicular to Tarzan's velocity, so does no work.  Since the gravitational energy is zero at the bottom point, the entire 1000 J of system mechanical energy must be converted to kinetic energy.  Now use the kinetic energy formula to calculate speed.

Summarizing: 1000 J gravitational energy before --> no work during --> 1000 J kinetic energy after.

2. Treating Tarzan as a single object:  Objects can have kinetic energy, but not potential energy.  So Tarzan begins with no kinetic energy.  During the process, two forces act on Tarzan: the force of the string, and the force of the earth.  The string force does no work on Tarzan because it is always perpendicular to Tarzan's velocity.  The earth does work on Tarzan equal to Tarzan's weight multiplied by the distance Tarzan travels parallel to the weight, or 1000 J.  This is positive work, because the force of gravity is parallel, not antiparallel, to the downward component of Tarzan's displacement.  At the bottom, then, we know 1000 J of work has been done on Tarzan; this must change Tarzan's kinetic energy by 1000 J, since we're not treating Tarzan as a system with internal structure.  Now use the kinetic formula to calculate speed.

Summarizing: 0 J total energy before --> 1000 J work done by gravity during -->  1000 J kinetic energy after.

This may sound like semantics to you.  It certainly does to me.  But I'm telling you, the new exam will ask students to explain energy conversion and conservation with direct reference to an appropriate object or system.*  We must teach the students to describe energy conversion in the kind of language I've included above.

* Look at learning objective 5.B.1.2 in the curriculum framework, for example.

Furthermore, the students are expected to be able to understand when the model of an object or system fails.*  What does that mean?  We calculated 6.3 m/s for Tarzan at the bottom.  A follow-up question might suggest that Tarzan was actually moving 5.0 m/s at the bottom, and experiments verify that air resistance has no measurable effect on Tarzan.  The question may ask for a resolution of this discrepancy.

* That's LO 5.B.2.1 for those of you with an education doctorate.

A reasonable explanation is that treating Tarzan as an object isn't reasonable.  He could be spinning; the "missing" mechanical energy is accounted for by his rotational kinetic energy.  Similar questions could be posed in which an object seems to have less energy than predicted by a point-object model, but some internal structure in motion accounts for the total mechanical energy budget.


5 comments:

  1. What about spring potential? Can springs, even when considered on their own (as a singular object), have elastic potential energy?

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  2. Technically, you only get potential energy if you consider a system consisting of a stretched/compressed spring AND an object attached to that spring. Potential energy exists due to an interaction between objects.

    One way to think about this is, if you're JUST talking about the spring alone, how could it convert its elastic potential energy to kinetic energy - kinetic energy of what? You've got to consider an object attached to the spring, or at least consider the end coil as a separate entity, to understand how the elastic potential energy converts.

    This is highly technical stuff. If your student is creating correct energy bar charts, is making correct predictions and calculations, then it's not worth worrying about whether it's "elastic potential energy of the spring" or "elastic potential energy of the spring-block system."

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  3. > The string force does no work on Tarzan because it is always perpendicular to Tarzan's velocity. The earth does work on Tarzan equal to Tarzan's weight multiplied by the distance Tarzan travels parallel to the weight, or 1000 J. This is positive work, because the force of gravity is parallel, not antiparallel, to the downward component of Tarzan's displacement.

    Hi! thanks so much for your content :) I'm struggling with systems so I have a couple of questions, but how do you know that the work done by the string is perpendicular to Tarzan's velocity? Is it because work = Fcos() and cos(90degrees) = 0? Would this be true (tension not doing any work) for all pendulums when the system is defined as the earth-attached mass system?
    Also, why wouldn't the force done by gravity be 0 as well, if gravity works straight down, and Tarzan moves horizontally?

    Thanks so much.

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  4. Hi, Unknown! You are exactly right as to why the string does no work on the Tarzan-Earth system.

    As to the work done by gravity... gravity is the force of the earth on Tarzan. That's *internal* to the Tarzan-Earth system! So it isn't included with work done by external forces. The work done by the earth is accounted for by the gravitational potential energy, which we're allowed to use because Earth is part of the system!

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  5. Thank you so much! That makes so much sense!

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