Energy is the ability to do work. The SI unit of energy is the Joule ($J$).
Work
Work is given by the dot product of the force and the displacement: $$ \begin{equation}\begin{aligned} Work=\vec{F}\cdot \vec{s}\\ \end{aligned}\end{equation} $$
Recall that the dot product of two vectors is the sum of the products of the corresponding components. Given the vectors $\vec{a}$ and $\vec{b}$: $$ \begin{equation}\begin{aligned} \vec{a}&=\begin{pmatrix}x_1\\ y_1\\ \end{pmatrix}\\ \vec{b}&=\begin{pmatrix}x_2\\ y_2\\ \end{pmatrix}\\ \end{aligned}\end{equation} $$ The dot product is $$\vec{a}\cdot \vec{b}=x_1x_2+y_1y_2$$
The dot product of two vector quantities always produces a scalar quantity. It tells us how much of one vector is in the direction of the other vector. Thus the work tells us how much of the displacement of the object is in the direction of the force applied to the object.
Work done on vs. work done by an object
If the displacement of the object when a force is applied to it is in the same direction as the force applied, then work is done on the object. Here, the dot product of the force and the displacement is positive.
Conversely, if the displacement occurs in the opposite direction, then work is done on the object. The dot product for such cases is negative.
Example Given the force $F$ and displacement $s$, determine the work done: $$ \begin{equation}\begin{aligned} \vec{F}&=\begin{pmatrix}6\\ 1\end{pmatrix}\\ \vec{s}&=\begin{pmatrix}2\\ 5\end{pmatrix}\\ \end{aligned}\end{equation} $$
Solution The work done is the dot product of the force and displacement: $$ \begin{equation}\begin{aligned} work&=\vec{F}\cdot \vec{s}\\ &=\begin{pmatrix}6\\ 1\end{pmatrix}\cdot \begin{pmatrix}2\\ 5\end{pmatrix}\\ &=6\times 2 + 1\times 5\\ &=17\\ \end{aligned}\end{equation} $$
Kinetic energy
The kinetic energy of a body is the energy possessed by a body by virtue of its motion. Thus it is dependent on the speed of an object: $$ \begin{equation}\begin{aligned} KE=\frac12 mv^2\\ \end{aligned}\end{equation} $$ For a change in kinetic energy: $$ \begin{equation}\begin{aligned} \Delta KE&=KE_f-KE_i\\ &=\frac12 mv_f^2-\frac12 mv_i^2\\ &=\frac12 m(v_f^2-v_i^2)\\ \end{aligned}\end{equation} $$
KE as work done
Kinetic energy can be seen as the work done by a force $F$ in order to increase the speed of an object of mass $m$ from rest ($u=0$) to a value $v$: $$ \begin{equation}\begin{aligned} Work&=\vec{F}\cdot \vec{s}\\ \end{aligned}\end{equation} $$
where: $$ \begin{equation}\begin{aligned} v^2&=u^2+2as\\ \therefore s&=\frac{v^2-u^2}{2a}\\ &=\frac{v^2-0^2}{2a}\\ s&=\frac{v^2}{2a}\\ \end{aligned}\end{equation} $$ and: $$ \begin{equation}\begin{aligned} F=ma\\ \end{aligned}\end{equation} $$
The resulting work is: $$ \begin{equation}\begin{aligned} Work&=\vec{F}\cdot \vec{s}\\ &=ma\times \frac{v^2}{2a}\\ E_K&=\frac{1}{2}mv^2\\ \end{aligned}\end{equation} $$
Potential energy
This is the energy possessed by a body by virtue of its state or position within a field. For gravitational potential energy, the value is dependent on the height, $h$: $$ \begin{equation}\begin{aligned} GPE=mgh\\ \end{aligned}\end{equation} $$
For a change in gravitational potential energy: $$ \begin{equation}\begin{aligned} \Delta GPE&=GPE_f-GPE_i\\ &=mgh_f-mgh_i\\ &=mg(h_f-h_i)\\ &=mg\Delta h\\ \end{aligned}\end{equation} $$
GPE as work done
The gravitational potential energy can be seen as the work done by the force of gravity ($W=mg$) in order to pull an object of mass $m$ vertically downwards through a distance of $h$: $$ \begin{equation}\begin{aligned} Work&=\vec{F}\cdot \vec{s}\\ &=mg\times h\\ E_P&=mgh\\ \end{aligned}\end{equation} $$
The Principle of Conservation of Energy
This states that energy is neither created nor destroyed, only changed from one state to another.
Given that the gravitational potential energy before an object is released to fall freely is equal to the kinetic energy at the bottom of its fall: $$ \begin{equation}\begin{aligned} \frac12 mv^2&=mgh\\ \frac12 v^2&=gh\\ v^2&=2gh\\ v&=\sqrt{2gh}\\ \end{aligned}\end{equation} $$
Thus if we know the height from which an object is dropped, we can predict the final velocity it has just before it hits the ground, assuming that there are no resistive forces acting.