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 \cdot \vec v</math>
+
::<math>= \frac{m}{2} 2 \vec \dot v \cdot \vec v  = \vec F_{\mbox{net}} \cdot \vec v</math>
  
 
or  
 
or  
  
−
<math>dT = \vec F \cdot \vec v dt =\vec F \cdot d\vec r \equiv d W</math>
+
<math>dT = \vec F_{\mbox{net}} \cdot \vec v dt =\vec F_{\mbox{net}} \cdot d\vec r</math>
  
 
or  
 
or  
  
−
:<math>\Delta T = \Delta W</math>
+
:<math>\Delta T =\int  \vec F_{\mbox{net}} \cdot d\vec r \equiv 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
 
;: The change in a particle's kinetic energy is equivalent to the work done by the net Force used to move the particle
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If we start with the form derived above
 
If we start with the form derived above
  
−
:<math>dT =\vec F \cdot d\vec r</math>
+
:<math>dT =\vec F_{\mbox{net}} \cdot d\vec r</math>
  
 
The change in the kinetic energy between two points and the corresponding work done as a result are
 
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
+
: <math>T_2-T_1 = \sum \vec F_{\mbox{net}} \cdot d\vec r = \int_1^2 \vec F_{\mbox{net}} \cdot d\vec r</math> in the limit of dr approaching zero
  
  
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And the kinetic energy decreases
 
And the kinetic energy decreases
 +
 +
 +
; also
 +
 +
remember
 +
 +
:<math>\vec F_{\mbox{net}} = \sum_{i=1}^n \vec {F}_i</math>
 +
 +
inserting this explicitly
 +
 +
: <math>T_2-T_1 =  \int_1^2 \sum_{i=1}^n \vec {F}_i \cdot d\vec r</math>
 +
 +
You can either do the summation first and then integrate or integrate and then do the sum
 +
 +
In practice you usually do calcualte the integral part first by calculating the work done by each force then add them all up.
 +
 +
  
 
===Problem 4.2===
 
===Problem 4.2===
 +
 +
Consider a force given by
 +
 +
:<math>\vec F = x^2 \hat i + 2xy \hat j</math>
 +
 +
a.)
 +
<math>W = \left ( \int_0^P \vec F \cdot dx \hat i + \int_0^P  \vec F \cdot dy \hat j\right )</math>
 +
<math> = \left ( \int_0^Q \vec F \cdot d \vec r + \int_Q^P \vec F \cdot d \vec r\right )</math>
 +
:<math>= \left ( \int_0^1 \vec F \cdot dx \hat i + \int_0^1  \vec F \cdot dy \hat j\right )</math>
 +
:<math>= \left ( \int_0^1 x^2 dx + \int_0^1  2ydy \right )</math>
 +
:<math>= \left ( \frac{1}{3} + 1 \right ) = \frac{4}{3} </math>
 +
  
  
  
 
[[Forest_UCM_Energy#KE_.26_Work]]
 
[[Forest_UCM_Energy#KE_.26_Work]]

Latest revision as of 18:48, 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{net}} \cdot \vec v[/math]

or

[math]dT = \vec F_{\mbox{net}} \cdot \vec v dt =\vec F_{\mbox{net}} \cdot d\vec r[/math]

or

[math]\Delta T =\int \vec F_{\mbox{net}} \cdot d\vec r \equiv 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_{\mbox{net}} \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_{\mbox{net}} \cdot d\vec r = \int_1^2 \vec F_{\mbox{net}} \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


also

remember

[math]\vec F_{\mbox{net}} = \sum_{i=1}^n \vec {F}_i[/math]

inserting this explicitly

[math]T_2-T_1 = \int_1^2 \sum_{i=1}^n \vec {F}_i \cdot d\vec r[/math]

You can either do the summation first and then integrate or integrate and then do the sum

In practice you usually do calcualte the integral part first by calculating the work done by each force then add them all up.


Problem 4.2

Consider a force given by

[math]\vec F = x^2 \hat i + 2xy \hat j[/math]

a.) [math]W = \left ( \int_0^P \vec F \cdot dx \hat i + \int_0^P \vec F \cdot dy \hat j\right )[/math] [math] = \left ( \int_0^Q \vec F \cdot d \vec r + \int_Q^P \vec F \cdot d \vec r\right )[/math]

[math]= \left ( \int_0^1 \vec F \cdot dx \hat i + \int_0^1 \vec F \cdot dy \hat j\right )[/math]
[math]= \left ( \int_0^1 x^2 dx + \int_0^1 2ydy \right )[/math]
[math]= \left ( \frac{1}{3} + 1 \right ) = \frac{4}{3} [/math]



Forest_UCM_Energy#KE_.26_Work

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