Work, Energy and Power

The energy of a system is a measure of its capacity to do work.

Kinetic energy is the energy a body has by virtue of its motion.

E_{trans}=\frac{1}{2}mv^{2}

Notice that translational kinetic energy is defined in terms of speed rather than velocity, ie it does not depend on direction.

Since mass is kg, and speed is ms-1, the unit for energy is kg x ms-1 x ms-1:

1 joule = i J = 1 kg m2s-2

The work done on a body by any force is the energy transferred to or from that body by the action of the force.

W=\Delta E_{trans}=\frac{1}{2}mv^{2}-\frac{1}{2}mu^{2}

Notice that work can be negative, ie causing the body to slow down.

The equations of uniformly accelerated motion tell us that:

v_{x}^{2}=u_{x}^{2}+2a_{x}s_{x}

and since

a_{x}=\frac{F_{x}}{m}

it follows that

v_{x}^{2}=u_{x}^{2}+2\frac{F_{x}}{m}s_{x}

Multiplying both sides by m and dividing by 2 we get:

\frac{1}{2}v_{x}^{2}-\frac{1}{2}u_{x}^{2}=F_{x}s_{x}

And therefore

W=F_{x}s_{x}

Note about units: The joule is kg m2s-2 and the N is 1 kg m s-2. So 1 joule is 1 N m.

 

The scalar product (dot product) of two vectors a and b is:

$latex \mathbf{a\cdot b}=ab\cos \theta $

An equivalent expression is:

$latex \mathbf{a\cdot b}=a_{x}b_{x}+a_{y}b_{y}+a_{z}b_{z}$

Which implies that:

$latex \cos \theta=\frac{a_{x}b_{x}+a_{y}b_{y}+a_{z}b_{z}}{ab}$

 

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