Why does kinetic energy E_k=½mv²?

Students and non-specialist teachers alike wonder: whence the half?

This post is intended to be a diagrammatic answer to this question using a Singapore Bar Model approach: so pedants, please avert your eyes.

I am indebted to Ben Rogers’ recent excellent post on showing momentum using the Bar Model approach for starting me thinking along these lines.

Part the First: How to get the *wrong* answer

Imagine pushing an object with a mass m with a constant force F so that it accelerates with a constant acceleration a so that covers a distance s in a time t. The object was initially at rest and ends up moving at velocity v.

(On the diagram, I’ve used the SUVAT dual coding conventions that I suggested in a previous post.)

So let’s consider the work done on the object by the force:

Step 1:    work done = force x distance moved in the direction of the force

Step 2:    W_d = F x s

But remember s = v x t so:

Step 3:    W_d = F x vt

And also remember that F = m x a so:

Step 4:    W_d = ma x vt

Also remember that a = change in velocity / time, so a = (v – 0) / t = v / t.

Step 5:        W_d = m (v / t) x vt

The ts cancel so:

Step 6:    W_d = mv²

Since this is the work done on the object by the force, it is equal to the energy transferred to the kinetic energy store of the object. In other words, it is the energy the object has gained because it is moving — its kinetic energy, no less: E_k = mv².

On a Singapore Bar Model diagram this can be represented as follows:

The kinetic energy is represented by the volume of the bar.

But wait: E_k=mv²!?!?

That’s just wrong: where did the half go?

Houston, we have a problem.

Part the Second: how to get the *right* answer

The problem lies with Step 3 above. We wrongly assumed that the object has a constant velocity over the whole of the distance s.

It doesn’t because it is accelerating: it starts off moving slowly and ends up moving at the maximum, final velocity v when it has travelled the total distance s.

So Step 3 should read:

But remember that s = (average velocity) x t.

Because the object is accelerating at a constant rate, the average velocity is (v + u) / 2 and since u = 0 then average velocity is v / 2.

Step 3:    W_d= F x (v / 2) t

And also remember that F = m x a so:

Step 4:    W_d= ma x (v / 2) t

Also remember that a = change in velocity / time, so a = (v – 0) / t = v / t.

Step 5:        W_d = m (v / t) x (v / 2) t

The ts cancel so:

Step 6:    W_d= ½mv²

Based on this, of course, E_k = ½mv²
(Phew! Houston, we no longer have a problem.)

Using the Bar Model representation, the volume of the bar which is above the blue plane represents the kinetic energy of an object of mass m moving at a velocity v.

The reason it is half the volume of the bar and not the full volume (as in the incorrect Part the First analysis) is because we are considering the work done by a constant force accelerating an object which is initially at rest; the velocity of the object increases gradually from zero as the force acts upon it. It therefore takes a longer time to cover the distance s than if it was moving at a constant velocity v from the very beginning.

So there we have it, E_k = ½mv² by a rather circuitous method.

But why go “all around the houses” in this manner? For exactly the same reason as we might choose to go by the path less travelled on some of our other journeys: quite simply, we might find that we enjoy the view.