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