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The relationship between the kinetic energy of a point particle and the work done on the point particle. This theorem is one way to arrive at a mathematical definition of work. |
Statement of the Theorem
If all the influences on a point particle are represented as works, the net work done by the forces produces a change in the kinetic energy of the particle according to:
| Latex |
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| Wiki Markup |
{table:border=1|frame=void|rules=cols|cellpadding=8|cellspacing=0} {tr:valign=top} {td:width=350|bgcolor=#F2F2F2} {live-template:Left Column} {td} {td} h1. The Work-Kinetic Energy Theorem {excerpt}The relationship between the [kinetic energy] of a [point particle] and the [work] done on the point particle. This theorem is one way to arrive at a mathematical definition of work.{excerpt} h3. Statement of the Theorem If all the influences on a [point particle] are represented as [works|work], the net work done by the forces produces a change in the kinetic energy of the particle according to: {latex}\begin{large}\[ \Delta K = W_{\rm net}\]\end{large}{latex} h3. Derivation of the Theorem From [ |
Derivation of the Theorem
From Newton's
...
...
...
for
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a
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point
...
particle,
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we
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know
| Latex |
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}\begin{large}\[ \vec{F}_{\rm net} = m\frac{d\vec{v}}{dt}\]\end{large}{latex} |
Now
...
suppose
...
that
...
the
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particle
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undergoes
...
an
...
infinitesimal
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displacement
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dr
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.
...
Since
...
we
...
want
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to
...
bring
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the
...
left
...
side
...
of
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the
...
equation
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into
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line
...
with
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the
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form
...
of
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the
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expression
...
for
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work,
...
we
...
take
...
the
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dot
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product
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of
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each
...
side
...
with
...
the
...
displacement:
| Latex |
|---|
}\begin{large}\[ \vec{F}_{\rm net}\cdot d\vec{r} = m\frac{d\vec{v}}{dt} \cdot d\vec{r} \]\end{large}{latex} |
Before
...
we
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can
...
integrate,
...
we
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make
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a
...
substitution.
...
Since
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v
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is
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the
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velocity
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of
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the
...
particle,
...
we
...
can
...
re-express
...
the
...
infinitesimal
...
displacement
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as:
| Latex |
|---|
}\begin{large}\[ d\vec{r} = \vec{v}dt\]\end{large}{latex} |
Making
...
this
...
substitution
...
on
...
the
...
right
...
hand
...
side
...
of
...
the
...
equation,
...
we
...
have:
| Latex |
|---|
}\begin{large}\[ \vec{F}_{\rm net}\cdot d\vec{r} = m\frac{d\vec{v}}{dt}\cdot \vec{v}\:dt = m\vec{v}\cdot d\vec{v} = m(v_{x}\;dv_{x} + v_{y}\;dv_{y}+v_{z}\;dv_{z})\]\end{large}{latex} |
We
...
can
...
now
...
integrate
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over
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the
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path:
| Latex |
|---|
}\begin{large}\[ \int_{\rm path} \vec{F}_{\rm net} \cdot d\vec{r} = \frac{1}{2}m(v_{x,f}^{2}-v_{x,i}^{2} + v_{y,f}^{2} - v_{y,i}^{2} + v_{z,f}^{2} - v_{z,i}^{2}) = \frac{1}{2}m(v_{f}^{2}-v_{i}^{2})\]\end{large}{latex} |
which
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is
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equivalent
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to
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the
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Work-Kinetic
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Energy
...
Theorem.
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