Now that we have a definition of work, we can apply the concept to kinematics. Just as force was related to acceleration through F = ma , so is work related to velocity through the Work-Energy Theorem.
It would be easy to simply state the theorem mathematically. However, an examination of how the theorem was generated gives us a greater understanding of the concepts underlying the equation. Because a complete derivation requires calculus, we shall derive the theorem in the one-dimensional case with a constant force.
Consider a particle acted upon by a force as it moves from x _{o} to x _{f} . Its velocity also increases from v _{o} to v _{f} . The net work on the particle is given by:
Using Newton's Second Law we can substitute for F:
Given uniform acceleration, v _{f} ^{2} - v _{I} ^{2} = 2a(x _{f} - x _{o}) . Substituting for a(x _{f} - x _{o}) into our work equation, we find that:
W _{net} = mv _{f} ^{2} - mv _{o} ^{2} |
As is evident by the title of the theorem we are deriving, our ultimate goal is to relate work and energy. This makes sense as both have the same units, and the application of a force over a distance can be seen as the use of energy to produce work. To complete the theorem we define kinetic energy as the energy of motion of a particle. Taking into consideration the equation derived just previously, we define the kinetic energy numerically as:
K = mv ^{2} |
Implying that
W _{net} = ΔK |
Though the full applicability of the Work-Energy theorem cannot be seen until we study the conservation of energy, we can use the theorem now to calculate the velocity of a particle given a known force at any position. This capability is useful, since it relates our derived concept of work back to simple kinematics. A further study of the concept of energy, however, will yield far greater uses for this important equation.