Difference between revisions of "Forest UCM Energy KEnWork"
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: <math>\frac{dT}{dt} = \frac{m}{2} \frac{d}{dt}v^2= \frac{m}{2} \frac{d}{dt}\vec v \cdot \vec v</math> | : <math>\frac{dT}{dt} = \frac{m}{2} \frac{d}{dt}v^2= \frac{m}{2} \frac{d}{dt}\vec v \cdot \vec v</math> | ||
::<math>= \frac{m}{2} \left (\vec \dot v \cdot \vec v + \vec v \cdot \vec \dot v \right )</math> | ::<math>= \frac{m}{2} \left (\vec \dot v \cdot \vec v + \vec v \cdot \vec \dot v \right )</math> | ||
− | ::<math>= \frac{m}{2} 2 \vec \dot v \cdot \vec v = \vec F_{\mbox{ | + | ::<math>= \frac{m}{2} 2 \vec \dot v \cdot \vec v = \vec F_{\mbox{tot}} \cdot \vec v</math> |
or | or |
Revision as of 02:48, 22 September 2014
Definition of KE
For a single particle of mass m moving with a velocity v, the kinetic energy is defined as
Work Energy Theorem
Derivation
Consider the Kinetic Energy's temporal rate of change assuming that the mass of the particle is constant
or
or
- The change in a particle's kinetic energy is equivalent to the work done by the net Force used to move the particle
Line Integral
If we start with the form derived above
The change in the kinetic energy between two points and the corresponding work done as a result are
- in the limit of dr approaching zero
- Negative Work?
Notice that if the Force is in the opposite direction of the displacement
Then negitive work is done
And the kinetic energy decreases