Let's first do some work and then we talk about the kinetic energy. Are you ready to do some work?

We consider a box on a frictionless horizontal surface where we apply a force for a particular displacement. The Figure 1 shows such arrangement. We apply a force on the box and we do work on it. Because there is no opposing force such as friction and let's imagine there are no other interfering forces.

Note that we applied a constant force (in this case net force because there are no other forces) \(F\) up to a displacement \(x\) in positive x-direction, now the work you did on the box is

\[W = F\,x \tag{1} \label{1}\]

Initially the box was at zero velocity (when it just started to moving), so \(v_0 = 0\) and the final velocity of the box is \(v\). The force is constant and the acceleration is constant which is \(a\). Using the appropriate kinematic equation, in this case \(v^2 = v_0^2 + 2ax\), you can write

\[ x = \frac{v^2}{2a}\]

Now substituting the value of \(x\) in the above equation you get

\[W = \frac{1}{2}m\,v^2\]

The above equation tells us that the work we did is stored in the motion of the box, and it is stored as the *energy* in the motion. The energy of a body associated with the motion of a body is called kinetic energy. Basically what happened here is that the work is converted to the kinetic energy. The work can be converted to different kinds of energies, so we need to separate them and we denote this kinetic energy by \(K\), and the kinetic energy is

\[K = \frac{1}{2}m\,v^2\]

Our kinetic energy has has speed with it. Why speed, not velocity? It's because it contains \(v^2\) and it is always positive and the sign of \(v\) has no meaning in determining the kinetic energy. So any object moving with speed \(v\) has kinetic energy \(1/2 m\,v^2\) regardless of the direction of motion.

In potential energy, you'll see that the work you do will be converted to different kind of energy called potential energy which is not actually associated with motion but position.