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{net}} \cdot \vec v</math>
+
::<math>= \frac{m}{2} 2 \vec \dot v \cdot \vec v  = \vec F_{\mbox{tot}} \cdot \vec v</math>
  
 
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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

[math]T \equiv \frac{1}{2} mv^2[/math]

Work Energy Theorem

Derivation

Consider the Kinetic Energy's temporal rate of change assuming that the mass of the particle is constant

[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} 2 \vec \dot v \cdot \vec v = \vec F_{\mbox{tot}} \cdot \vec v[/math]

or

[math]dT = \vec F \cdot \vec v dt =\vec F \cdot d\vec r \equiv d W[/math]

or

[math]\Delta T = \Delta W[/math]
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

[math]dT =\vec F \cdot d\vec r[/math]

The change in the kinetic energy between two points and the corresponding work done as a result are

[math]T_2-T_1 = \sum \vec F \cdot d\vec r = \int_1^2 \vec F \cdot d\vec r[/math] 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

Problem 4.2

Forest_UCM_Energy#KE_.26_Work