Forest UCM Energy KEnWork

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 \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_{\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

remember

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

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