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### Integration and the Concept of Work

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Date: 3/19/96 at 20:5:42
From: J. DeMeo
Subject: (no subject)

In calculus class we are studying work.  If asked to find the work
done by a varying force F(x) = 4x^3 from -4 to +4, using
integration the work done equals 0.  Mathematically I understand
why that is, yet how can it be physically that no work is done
when applying this force over a distance of 8 units?  The calculus
book says the answer really is zero, but how do you convince
someone that this makes sense?  Is there a way to make an analogy
in the real (physical) world?

Thanks
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Date: 3/21/96 at 13:17:59
From: Doctor Aaron
Subject: Re: (no subject)

Hi,

The concept of work in physics is a little different from everyday
usage, so it takes a little getting used to.  One way to look at
it is to break the integral up into 2 regions:

4          0       4
::         ::      ::
W= -:  F   =  -: F +  -: F. If you evaluate both of these integrals, you
::         ::      ::     will get 256 + -256 = 0.
-4         -4       0

The physical interpretation of this is that, while the object is
on the left (negative) side of the x-axis, the force is negative,
so the object is being pushed in a negative direction.  By a
physicist's definition, work is being done (256 units worth of
it).  However, when the object is on the right (positive) side of
the axis, it is being pulled away from the origin so negative work
is being done (again, 256 units worth).  This idea of negative
work doesn't make sense if you think of work as stuff that makes
you tired, but in physics it is very useful.

There is one other thing in my explanation that should confuse
you.  I said that the object gets pushed away while it is on the
negative side of the axis.  You might thing that it should never
get to the positive side.  Now think of throwing the object from
-4 on the x-axis, the force (4x^3) will be negative until it gets
to the origin.  This force will slow down the object a great deal,
but once it gets to the positive side of the axis, the forces will
push the object in the positive direction.  It turns out, because
of the anti-symmetry about the origin of the way that the force
acts, that once the object gets to 4 on the x-axis, it will have
exactly the same speed as it did at -4, when you threw it
(provided you threw it fast enough that the object made it to the
origin without turning around).

If it is helpful, you can think of a configuration of planets that
produce a gravitational force field, or a configuration of charged
objects that produce an electro-magnetic force field, such that
the force is 4x^3 on the x-axis between -4 and 4.  If a meteor or
charged object flew through this force field, it would have the
same velocity at -4 and +4 on the x-axis.

I hope that this is helpful, if you need further clarification,
don't hesitate to write back.

-Doctor Aaron,  The Math Forum

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Associated Topics:
High School Physics/Chemistry

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