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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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\begin{large}\[ \Delta K = W_{\rm net}\]\end{large} |
Derivation of the Theorem
From Newton's 2nd Law for a point particle, we know
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\begin{large}\[ \vec{F}_{\rm net} = m\frac{d\vec{v}}{dt}\]\end{large} |
Now suppose that the particle undergoes an infinitesimal displacement dr. Since we want to bring the left side of the equation into line with the form of the expression for work, we take the dot product of each side with the 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} |
Before we can integrate, we make a substitution. Since v is the velocity of the particle, we can re-express the infinitesimal displacement as:
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\begin{large}\[ d\vec{r} = \vec{v}dt\]\end{large} |
Making this substitution on the right hand side of the equation, we 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} |
We can now integrate over the 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} |
which is equivalent to the Work-Kinetic Energy Theorem.