Although, energy and power are interrelated concepts, they possess distinct identities of their own. Consider the relatively simple task of inflating a balloon. Blowing a small puff of air into a balloon over a short time-interval will cause the balloon to expand slightly before recoiling to its previous state. Breathing more forcefully into a balloon will cause it to expand more-so, but if the rate at which energy is deposited into this system is too small, the balloon will continue to recoil to its original disposition shortly afterward; therefore, successful inflation requires a threshold energy input that is deposited at a rate that doesn’t allow it to recoil. Energy is defined as the ability of a force to do work through some distance, and the SI unit for energy is the joule ( J ). The rate at which energy is deposited within a system is called power ( P ), and its SI unit is the watt ( W ):
P = ( J / s )
Let’s now consider a circumstance in which a skydiver is in freefall. In the beginning, the skydiver will accelerate toward the earth under the influence of an unbalanced force:
F = ma
Recall that Newton’s Third Law of Motion states that every force is accompanied by an equal and opposite force. As the skydiver exerts more force upon the molecules in the atmosphere, the atmosphere pushes back with more force until freefall at a constant velocity is established. An object that moves with a constant velocity possesses a measure of inertia in accordance with its massiveness:
p = mv
As the skydiver falls a given distance, she performs work upon the atmosphere, and in turn, the atmosphere performs work upon her. The work the atmosphere must do to establish a constant velocity is quantified as follows:
W = Fd
And, of course, force is a product of mass and acceleration:
F = ma
Somehow or another, all of the equations above apply to the scenario, but how? An analytic premise can be established via mathematical substitution:
W = Fd
W = ( ma )( d )
Let’s now use the equation for momentum to further expand the equation at hand:
p = mv
m = ( p / v )
F = ma = ( p / v )( a ) = [ ( pa ) / v ]
Fv = pa = ( mv )( a ) = ( ma )( v )
The lattermost equation is somewhat ambiguous, because it contains the velocity expression on both sides; however, velocity is the distance the skydiver will fall in a given unit of time ( t ):
v = ( m / s ) = ( distance / time )
And therefore,
Fv = ( F )( distance / time ) = ( Fd / s ) = ( J / s ) = P
Fortunately, power derivations in the field of electronics ( although different ) can be obtained from mathematical manipulation of the voltage equation:
V = IR
This exercise, however, should only be continued after the aforementioned conceptual analysis is thoroughly understood.
RENAISSANCE PHYSICS 