📂Classical Mechanics

Day, Work-Energy Theorem

Definition

Force $\mathbf{F}$ moves a body from point $\mathbf{a}$ to $\mathbf{b}$ along a path $C$; the following path integral (line integral) is called the work done by that force, denoted work.

$$ W = \int_{C} \mathbf{F} \cdot \mathrm{d} \mathbf{r} \tag{1} $$

Explanation

The above integral equals the area under the force–displacement graph. The reason it is defined using the dot product (inner product) is to count only the component of the force that contributes in the actual direction of motion. If the force is constant (magnitude and direction do not change) and the object undergoes straight-line motion, $(1)$ simplifies as follows. For the displacement vector $\mathbf{s} = \mathbf{b} - \mathbf{a}$,

$$ W= \mathbf{F} \cdot \mathbf{s} $$

Units

In the SI system, the unit of work is the same as that of energy: the joule.

$$ 1\ \mathrm{J} = 1\ \mathrm{N} \cdot \mathrm{m} = 1\ \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}^{2} $$

As a side note, because the joule is named after a person its symbol is written as the uppercase $\mathrm{J}$, but when referring to the unit in text it is written lowercase as joule.

Work–Energy Theorem¹

The net work done on an object by the net force equals the change in its kinetic energy.

$$ W=\Delta T $$

Proof

Actually, it is not really a proof: the following relation is derived from the definition of work (see ../507), and from this the kinetic energy is defined.

$$ W_{\mathbf{a}\mathbf{b}} = \int_{\mathbf{a}}^{\mathbf{b}} \mathbf{F} \cdot \mathrm{d}\mathbf{r} = \dfrac{1}{2} m v_{\mathbf{b}}^{2} - \dfrac{1}{2} m v_{\mathbf{a}}^{2} $$

■