Re: Spring potential energy without Work (sort of)

[John Clement <clement@HAL-PC.ORG>]

Another method of deriving results is used in MOP (Minds on Physics).  They
have introduced the idea of calculating the area under the curve (backdoor
calculus?).  This can be used to calculate the work for an ideal spring
using the F-x curve.  This is used in a number of places to derive results,
but it is often done in an informal manner by having students calculate
numbers.  The equation X=X0 + V0 t + 1/2 a t^2 is actually derived
symbolically using graphs and the idea that the area under the velocity time
curve for constant acceleration is the displacement.  they do not use all of
the SVT equations, but rather have the students do problems using strobe
diagrams, graphs, and once in a while equations.  Their method promotes
concept based problem solving, and actually allows students in an algebra
based course to solve very complex problems.  According to one study (on
their web site) their methods achieve expert like problem solving ability in
students.

John M. Clement
Houston, TX

> > It seems to me that it involves as much calculus as deriving
> > (1/2 mv^2).
>
> Exactly!!
>
> > This is usually done (in an algebra course) by stating that if v
> > increases linearly with time then v_avg = (v_i + v_f)/2.
>  >If
> > this hand-
> > waving works, then we might as well say that if F increases linearly
> > with distance then F_avg = (F_i + F_f)/2, no?*
>
> I see your point.
>
> > Is this hand-waving somehow worse than the hand-waving
> > performed to get
> > v_f^2 = v_i^2 + 2 a Delta x?
>
> Doesn't require handwaving at all once you have the velocity and
> position as
> functions of time for constant acceleration.
>
> The hand waving occurs for obtaining position as a function of time, a
> natural function to want if you first have v(t) for a constant
> acceleration.
> The delta V^2 equation is easily motivated by saying lets eliminate t from
> the other two equations.
>
> Furthermore, average velocity doesn't require handwaving, algebra and
> geometry alone serve to derive this for constant acceleration; at least if
> you believe geometry implies that the midpoint of a linear line is its
> average value, (and don't call that handwaving).  But we're
> getting into the
> realm of taste here, IMO the result is less natural for Delta x^2; in the
> sense that I think it requires knowing the desired answer in a
> slightly more
> profound way than the Delta v^2.
>
> I readily admit that formally the arguments are identical.
>
> > *[since F = kx, then
> > F_avg*Delta x = k (x_i + x_f)*(x_f - x_i)/2 = 1/2 k Delta (x^2).]
>
> Good!  We don't need calculus for the linear spring! And I suppose this
> answers, perhaps not in pleasing way, Ludwik's question.
>
> Of course, I'll next say, let's consider a non-linear spring, . . .
>
> Joel R