Kinetic energy
Energy can be understood intuitively as ‘the ability of an object to effect change on its environment’ or ’the ability of an object or system to do work’. We shall see that different types of energy exist.
The energy associated with motion is called kinetic energy.
Consider a car in motion that smashes into another car. The kinetic energy of the car in motion can be measured by the extent of the damage done. We feel that the faster the car goes, the greater the damage, so we can conclude that kinetic energy increases with the speed of the object. We can also argue that the heavier the car, the greater the damage, so kinetic energy must be proportional to the mass of the object.
Definition
Kinetic energy is half the product of the mass and the speed of the object squared:
\[E_{\mathrm{K}} := \frac{1}{2} m v^2.\]
Kinetic energy depends on the speed and not on the velocity: even if the direction of the velocity changes, the kinetic energy stays constant if the magnitude of the velocity does not change.
Kinetic energy, as energy in general, is a scalar quantity.
The SI unit of the kinetic energy, and energy in general, is the same as that of work, the joule:
\[\left[E_{\mathrm{K}}\right] = \left[E\right] = 1\,\mathrm{N} \cdot \mathrm{m} = 1\,\mathrm{kg} \cdot \frac{\mathrm{m}^2}{\mathrm{s}^2} = 1\,\mathrm{J}.\]
Why \(\frac12 mv^2\)?
First, consider an example. We observe an object of mass \(m = 1\,\mathrm{kg}\) in free fall which started from rest and record its speed, displacement, kinetic energy as well as the work done on it by gravity up to the current position in each second. Free fall is linear motion with constant acceleration with a starting speed of \(v_0 = 0\) and an acceleration of \(a = g,\) so
\[v(t) = g t,\]
and
\[\Delta h = \frac{g}{2} t^2.\]
The force of gravity \(F = m g\) acting on the object is constant, so in a displacement \(\Delta h,\) it does \(W = m g \Delta h\) work on the object. The values are the following:
We can see that if we define the kinetic energy as \(E_{\mathrm{K}} = \frac12 m v^2,\), its value will be equal to the work done on the object by gravity.
Now let us see something more general. For linear motion with constant acceleration, we can use the alternative displacement formula
\[2 a \Delta x = v^2 - v_0^2.\]
We can express the acceleration in terms of the net force using Newton‘s second law:
\[m a = \sum{F} \ \Rightarrow \ a = \frac{\sum{F}}{m}.\]
Substituting this into the displacement formula, we get
\[2 \left(\frac{\sum{F}}{m}\right) \cdot \Delta x = v^2 - v_0^2,\]
\[\sum{F} \cdot \Delta x = \frac12 m v^2 - \frac12 mv_0^2,\]
\[\sum{W} = \frac12 m v^2 - \frac12 m v_0^2 = \Delta E_{\mathrm{K}}.\]
What does it mean? It means that if we define the kinetic energy as \(\frac12 m v^2,\) its change \(\Delta E_{\mathrm{K}}\) in the process of linear motion with constant acceleration will be equal to the net work \(\sum W\) done on the object. This observation is called the work–kinetic energy theorem, and it holds true for any type of motion, not just linear motion with constant acceleration.