As you read earlier, the change in the total energy of a system is equal to the work done on that system by its surroundings. In these notes, you will read about the formal definition of work, which is the transfer of mechanical energy, and a mathematical idea that underpins work - the dot product.

The work that is done by a force is the scalar product (or dot product) of that force and the displacement.

$$W = \vec{F}\cdot\Delta\vec{r} = F_x dx + F_y dy + F_z dz$$

The dot product is one way that two vectors are “multiplied.” It is the sum of the product of each pair of components. This dot product is related to the angle that the force makes with the displacement. Essentially, the dot product will “pick out” the component of one vector that is parallel to another vector.

A point particle moves through a distance $d$ while a force $F$ is applied at an angle $\theta$ relative to the displacement.

Consider a point particle that moves through a displacement $\Delta \vec{r}$ while it experiences a force $\vec{F}$ at an angle $\theta$ relative to the displacement. The work that done on the particle by this force is,

$$W = \vec{F}\cdot\Delta\vec{r} = F_x \Delta x + F_y \Delta y + F_z \Delta z = F\cos\theta \Delta r = F\cos\theta d$$

where the last step considers that the only non-zero part of the dot product is the x bit. That is, the displacement is in the x-direction ($\Delta r$ = d), and the component of the force in that direction is $F\cos\theta$. In general, the work calculation picks out the piece of the force that is parallel to the displacement – that interaction is what increases the energy of the system.

$$W = \vec{F}\cdot\Delta\vec{r} = F_{\parallel}\Delta r$$

The units of work can be determined by the product of the units of its constituent bits.

$$Work = (Force)*(distance) = (Newtons)*(meters) = Nm = Joule$$

The units of work is a Joule named after