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Work-Energy theorem


In the 1840's James Prescott Joule set out to verify that gravitational potential energy converts to heat in a consistent way.

One of the setups used by James Prescott Joule

Among the setups used by Joule was one of the type shown in the image. A weight is attached to a rope, the rope runs over a pulley. At the start the rope is coiled around a vertical axle that makes paddles turn in water; the water is inside a calorimeter.

The paddles are churning the water and the effect of this churning is that the temperature of the water rises. Joule confirmed that double the height gives double the temperature change. That is, Joule confirmed that the relation is linear: if the force is uniform then force multiplied by distance gives the amount of energy.

This exemplifies that it is very useful to have an equation that expresses force acting over distance.

In the case of the paddle wheel setup used by Joule the friction limits the rate at which the weight descends. Other experiments corroborate that the amount of potential energy release on descent does not depend on the velocity of descent. If an object falls without friction the velocity increases linear with time. That corroborates that for free fall also the potential energy release is linear with height.

The Work-Energy theorem has been recognized for centuries, but for a long time it was not yet stated in its current form. So the Work-Energy theorem is both old and new; a history of twists and turns. For the rest of this exposition that history will be ignored, making it seem as if the theorem was always clearly recognized.


In the following exposition the Work-Energy theorem is derived twice. First in the way it could have been derived at the earliest opportunity. This derivation can also be seen as an exercise in economy; it uses the barest minimum of mathematical tools that will accomplish the job. The value of a minimal derivation is this: by removing everything that is *not* necessary it is demonstrated what the essence of the theorem is.

We have the following two expressions to describe motion of a point that is subject to uniform acceleration

$$ v = v_0 + at \qquad (1.1) $$

$$ s = s_0 + v_0t + \frac{1}{2}at^2 \qquad (1.2) $$

(1.1) is the derivative of (1.2), in that sense the two are not independent, but it's still possible to take advantage of the fact that there are two equations there.

If an object is subject to uniform acceleration $a$ over a distancs $(s - s_0)$ then how much change of velocity will that cause?

So, we want to start with acceleration and distance traveled as a given, and obtain velocity. That means we need to work towards an expression that contains position, acceleration and velocity, but not time.

We obtain an expression for '$t$' from (1.1), and substitute that into (1.2). After that substitution: while time is implicitly still present (velocity and acceleration are time derivatives), there is no explicit time.

$$ s = s_0 + v_0\frac{(v-v_0)}{a} + \tfrac{1}{2} a \frac{{(v-v_0)}^2}{a^2} \qquad (1.3) $$

We multiply everything out, and move terms for position and acceleration to the left.

$$ a(s-s_0) = vv_0 - {v_0}^2 + \tfrac{1}{2}v^2 - vv_0 + \tfrac{1}{2} {v_0}^2 \qquad (1.4) $$

Quite a few terms drop away against each other, and the result is:

$$ a(s - s_0) = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \qquad (1.5) $$

(1.5) looks very different from (1.1) and (1.2), but it is mathematically equivalent.

(1.5) is also known as Torricelli's equation

Next we combine (1.5) with $F=ma$

$$ F(s - s_0) = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \qquad (1.6) $$

Repeating the three equations that are sufficient to define the Work-Energy theorem:

$$ v = v_0 + at \qquad (1.1) $$

$$ s = s_0 + v_0t + \frac{1}{2}at^2 \qquad (1.2) $$

$$ F=ma $$

$F=ma$ and the work-energy theorem are very closely connected indeed. By taking the derivative with respect to position $F=ma$ is recovered immediately.

The following derivation is more abstract. This derivation is from the outset designed to accomodate an arbitrary acceleration profile.

In the course of the derivation the following two relations will be used:

$$ ds = v \ dt \qquad (2.1) $$

$$ dv = \frac{dv}{dt}dt \qquad (2.2) $$

The integral for acceleration from a starting point $s_0$ to a final point $s$.

$$ \int_{s_0}^s a \ ds \qquad (2.3) $$

Use (2.1) to change the differential from *ds* to *dt*. Since the differential is changed the limits change accordingly.

$$ \int_{t_0}^t a \ v \ dt \qquad (2.4) $$

Rearrange the order, and write the acceleration $a$ as $\tfrac{dv}{dt}$

$$ \int_{t_0}^t v \ \frac{dv}{dt} \ dt \qquad (2.5) $$

Use (2.2) for a second change of differential, again the limits change accordingly.

$$ \int_{v_0}^v v \ dv \qquad (2.6) $$

Putting everything together:

$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \qquad (2.7) $$

Combining with $F=ma$ gives the work-energy theorem

$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \qquad (2.8) $$

This form of the work-energy theorem emphasizes that in order to define a potential energy an integral must be defined.

Gravity is an example of a force that satisfies all the conditions in order for a potential energy to be defined:
- The difference in potential is independent of the way you go from one height to another.
- The acceleration caused by gravity is independent of any preexisting velocity, you just get more accumulation of velocity.

An obvious example of a class of types of force that do not satisfy the condition is any form of friction.


The motto in this section is: the higher the level of abstraction, the better.

Why is it interesting?

As stated in the lead up to the derivation: the Work-Energy theorem is both old and new.

The Work-Energy theorem is the basis of the formulation of mechanics that was developed by Joseph Louis Lagrange, called 'Lagrangian mechanics'. But at the time the Work-Energy theorem did not yet exist in its current form. The modern formulation of the Work-Energy theorem is not present in Lagrange's work. The form that Lagrange used is called d'Alembert's virtual work.

Kinetic energy and Lagrangian mechanics

Motion of objects is represented with coordinates of a coordinate system. In a system of cartesian coordinates a velocity vector can be decomposed in components along the axes of the cartesian coordinate system. In order to represent momentum in a cartesian coordinate system with three dimensions the momentum is decomposed along the three axes of the coordinate system. Likewise the force is decomposed along the three axes. Newton's second law is valid for each component of motion.

$ \vec{F_x} = m \vec{a_x} $
$ \vec{F_y} = m \vec{a_y} $
$ \vec{F_z} = m \vec{a_z} $

Kinetic energy is proportional to the square of velocity. This means that the sum of the component kinetic energies (along the coordinate axes) is equal to the kinetic energy of the undecomposed velocity (pythagoras' theorem). This enables the following: in Lagrangian mechanics the direction of the velocity vector is discarded. The kinetic energy is treated as a scalar.

The directional information of the velocity vector can be discarded because that information is still available in the expression for the potential energy. The potential energe is the integral of the force over distance. When the calculation is for motion in three dimensions of space then the expression for the potential energy has three components, one for each spatial dimension.

Kinetic energy can be treated as a scalar without loss of expressive power; that is what Joseph Louis Lagrange capitalized on when he developed Lagrangian mechanics.

The second innovation from Lagrange was systematic use of generalized coordinates. The importance of being able to use generalized coordinates cannot be overstated.

theorems/work-energy_theorem.txt · Last modified: 2021/04/25 12:47 by cleonis