In physics, we take measurable quantities from the real world, and attempt to find meaningful relationships between them. A basic example of this would be the physical ideal of force. Force applied to an object changes the motion of an object. Here’s the deal though, at a basic level

and while we can put a physical interpretation to this arithmetical definition, at the end of the day force is simply “mass times acceleration.” The SI unit of force is a newton, which is defined to be

In a similar way, the idea of kinetic energy, is “energy” objects have from motion. It is defined by the formula

The SI unit of energy is a joule, which is defined to be

To get a feel for the “size” of a joule, consider this: if an apple has a mass of \(0.1\mathrm {kg}\) and it is dropped from a height of \(1\mathrm {m}\), then approximately \(1\) joule of energy is released when it hits the ground. Let’s see if we can explain why this is true.

Finally work is defined to be accumulated force over a distance. Note, there must be some force in the direction (or opposite direction) that the object is moving for it to be considered work.

We can write the definition of work in the language of calculus as,

The SI unit of work is also a joule. To help understand this, \(1\) joule is approximately how much work is done when you raise an apple one meter.

Let’s again see why this is true.

Now we have a question:

Why do work and kinetic energy have the same units?

One way to answer this is via the Work-Energy Theorem.

Work-Energy Theorem Suppose that an object of mass \(m\) is moving along a straight line. If \(s_0\) and \(s_1\) are the the starting and ending positions, \(v_0\) and \(v_1\) are the the starting and ending velocities, and \(F(s)\) is the force acting on the object for any given position, then

\[ W = \int _{s_0}^{s_1} F(s) ds = \frac {m\cdot v_1^2}{2} - \frac {m\cdot v_0^2}{2}. \]

First we need to get all of our symbolism out in the open. Let:

• \(s(t)\) represent position with respect to time,
• \(v(t)\) represent velocity with respect to time,
• \(a(s)\) represent acceleration with respect to position,
• \(t_0\) represent the starting time,
• \(t_1\) represent the ending time,

then we also have that

• \(s(t_0)\) represents the starting position, \(s_0\),
• \(s(t_1)\) represents the ending position, \(s_1\),
• \(v(t_0)\) represents the starting velocity, \(v_0\),
• \(v(t_1)\) represents the ending velocity, \(v_1\).

Now write with me,

\[ W = \int _{s_0}^{s_1} F(s) ds = \int _{s(t_0)}^{s(t_1)} F(s) ds \]

here we are working with functions of distance. We will use the substitution formula,

transforming from right to left, to see that

\[ \int _{s(t_0)}^{s(t_1)} F(s) ds = \int _{t_0}^{t_1} F(s(t)) s'(t) dt \]

and we are now working with functions of time. Since \(s'(t) = v(t)\), we may write

\[ \int _{t_0}^{t_1} F(s(t)) s'(t) dt = \int _{t_0}^{t_1} F(s(t)) v(t) dt \]

and now remember that \(F=m\cdot a\), so

\[ \int _{t_0}^{t_1} F(s(t)) v(t) dt = \int _{t_0}^{t_1} m\cdot a(s(t)) v(t) dt. \]

However, \(a(s(t)) = v'(t)\), so rearranging we have,

\[ \int _{t_0}^{t_1} m\cdot a(s(t)) v(t) dt = m\cdot \int _{t_0}^{t_1} v(t) v'(t) dt. \]

Now we apply the substitution formula again, this time we will transform left to right

and so we see

\[ m\cdot \int _{t_0}^{t_1} v(t) v'(t) dt = m\cdot \int _{v(t_0)}^{v(t_1)} v dv \]

and we are working with functions of velocity. At last, setting \(v(t_0) = v_0\) and \(v(t_1) = v_1\), we can evaluate this integral, \begin{align*} m \cdot \int _{v_0}^{v_1} v dv &= m \cdot \bigg [ \frac {v^2}{2} \bigg ]_{v_0}^{v_1}\\ &=\frac {m\cdot v_1^2}{2} - \frac {m\cdot v_0^2}{2}. \end{align*}

The Work-Energy theorem says that:

This could be interpreted as:

The accumulated force over distance is the change in kinetic energy.

Moreover, this answers our initial question of why work and kinetic energy have the same units. In essence, energy powers work.