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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:
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| Wiki Markup |
{excerpt:hidden=true}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} h4. 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} h4. Derivation of the Theorem From [ |
Derivation of the Theorem
From Newton's
...
...
...
for
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a
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point
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particle,
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we
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know
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}\begin{large}\[ \vec{F}_{\rm net} = m\frac{d\vec{v}}{dt}\]\end{large}{latex} |
Now
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suppose
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that
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the
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particle
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undergoes
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an
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infinitesimal
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displacement
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dr
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.
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Since
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we
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want
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to
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bring
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the
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left
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side
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of
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the
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equation
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into
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line
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with
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the
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form
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of
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the
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expression
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for
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work,
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we
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take
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the
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dot
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product
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of
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each
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side
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with
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the
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displacement:
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}\begin{large}\[ \vec{F}_{\rm net}\cdot d\vec{r} = m\frac{d\vec{v}}{dt} \cdot d\vec{r} \]\end{large}{latex} |
Before
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we
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can
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integrate,
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we
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make
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a
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substitution.
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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
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particle,
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we
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can
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re-express
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the
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infinitesimal
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displacement
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as:
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|---|
}\begin{large}\[ d\vec{r} = \vec{v}dt\]\end{large}{latex} |
Making
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this
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substitution
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on
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the
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right
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hand
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side
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of
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the
...
equation,
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we
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have:
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}\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
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can
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now
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integrate
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over
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the
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path:
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}\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.