Re: Spring potential energy without Work (sort of)

["RAUBER, JOEL" <JOEL_RAUBER@SDSTATE.EDU>]

> 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