Work-Kinetic Energy Theorem

Recall that for constant acceleration in 1 dimension:

$$v_f^2 - v_i^2 = 2 a \Delta x$$ (3.1)

where $\Delta x$ is the displacement in the direction of the acceleration.

If we multiply by $m$ (the mass of the object) and move the annoying 2 over to the other side, we can make the constant $a$ into a constant force $F_x = ma$:

$$(ma)\Delta x = \frac{1}{2}m v_f^2 - \frac{1}{2}m v_i^2$$ (3.2)

$$F_x \Delta x = \frac{1}{2}m v_f^2 - \frac{1}{2}m v_i^2$$ (3.3)

We now define the work done by the force $F_x$ on the mass $m$ to be:

$$\Delta W = F_x \Delta x.$$ (3.4)

The work can be positive or negative. It has SI units of N-m = Joules = J.

Work is a form of energy. When we do work on an object with mass $m$, we change the quantity

$$E_k = \frac{1}{2}m v^2$$ (3.5)

which is its energy of motion or kinetic energy. Note that this is a relative quantity - it depends upon the inertial frame in which it is measured. For example, the kinetic energy of a ball being tossed in a car is small relative to the car, but may be quite large relative to the ground flying by the car at 100 kph and extremely large relative to the sun as the earth whips around it.

Physics is the study of dynamics and forces that change the velocity (and hence the kinetic energy). The work done on a mass changes its kinetic energy by a certain amount independent of what it was to begin with. We thus start to see that dynamics (that is, our analysis of the effects of the forces involved) will not depend upon any constant we add to the energies being studied, as the depend only on the difference in energy between one state and another.

We now can formally state the work-energy theorem in the first form: The work done on a mass by the total force acting on it is equal to the change in its kinetic energy, or:

$$\Delta W = F_x \Delta x = \frac{1}{2}m v_f^2 - \frac{1}{2}m v_i^2 = \Delta E_k$$ (3.6)

So far, we have proven this only for constant $F = ma$ and one dimensional motion. However, we can easily show it to be true more generally.

In this figure we see that we can break up a non-constant force acting through some distance into small intervals. In each interval the force is approximately constant. We thus get:

$$\Delta W_T = F_1 (x_1 - x_0) + F_2 (x_2 - x_1) + \ldots$$ (3.7)

$$= (\frac{1}{2}m v^2(x_1) - \frac{1}{2}m v^2(x_0)) + (\frac{1}{2}m v^2(x_2) - \frac{1}{2}m v^2(x_1)) + \ldots$$ (3.8)

$$= \frac{1}{2}m v^2_f - \frac{1}{2}m v^2_i$$ (3.9)

where the internal changes in kinetic energy at the end of each interval but the first and last cancel. Finally, we let $\Delta x$ go to zero in the usual way, and replace summation by integration. Thus:

$$\Delta W = \lim_{\Delta x \to 0} \sum_i F_i \Delta x = \int F dx = \Delta E_k$$ (3.10)

and we have generalized the theorem to include non-constant forces.

Finally, we can also show this to be true for each component of a force (and the kinetic energy associated with that component only). Because $v^2 = \vec{v} \cdot \vec{v} = v_x^2 + v_y^2 + v_z^2$, we can readily show that:

$$\Delta W = \vec{F} \cdot \Delta \vec{x} \to \int \vec{F} \cdot d\vec{x} = \Delta E_k$$ (3.11)

which is now our fully generalized work energy theorem in three dimensions.

Note well: Energy is a scalar and hence a bit easier to treat than vector quantities like forces.

Note well: Normal forces (perpendicular to the direction of motion) do no work:

$$\Delta W = \vec{F_\perp} \cdot \Delta \vec{x} = 0.$$ (3.12)

This really simplifies problem solving, as we shall see.