9.2.3 Extending the Model

Our initial model has been of a single particle acted on by a constant force parallel to the displacement. We can easily make some straightforward extensions of this model to slightly more complex—and interesting—situations:

Force perpendicular to the displacement: A force parallel to a particle’s displacement causes the particle to speed up or slow down, changing its energy. But a force perpendicular to the displacement does not change the particle’s speed; it is neither speeding up nor slowing down. Its energy is not changing, so no work is being done on it. A force perpendicular to the displacement does no work.
Multiple forces: If multiple forces act on a system, their works add. That is, $Δ K =$ $W_{tot}$ , where the total work done is

$W_{tot} = W_{1} + W_{2} + W_{3} + \dots$ (9.11)
Multiparticle systems: If a system has multiple independent particles, which don’t interact, the system’s energy is the total kinetic energy of all the particles:

$E_{sys} = K_{tot} = K_{1} + K_{2} + K_{3} + \dots$ (9.12)

$K_{tot}$ is truly a system energy, not the energy of any one particle. How does $K_{tot}$ change when work is done? You can see from its definition that $Δ K_{tot}$ is the sum of all the individual kinetic-energy changes, and each of those changes is the work done on that particular particle. Thus

Δ K_{tot} = W_{tot}

(9.13)

where now $W_{tot}$ is the total work done on all the particles in the system.

NOTE You might expect $W_{tot} = (F_{net})_{s} Δ s$ , where $\vec{F}_{net}$ is the net force on the system. This is true for a one-particle system (if all the forces are constant), but in general it is not true for a multiparticle system because each particle undergoes a different displacement. You must find the work done on each particle, then sum those to find the total work done on the system.