# Category Archives: Bar Model

## Why does kinetic energy = 1/2mv^2?

Why does kinetic energy Ek=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= F x s

But remember s = v x t so:

Step 3:    W= F x vt

And also remember that F = m x a so:

Step 4:    W= ma x vt

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

Step 5:        Wd = (v / t) x vt

The ts cancel so:

Step 6:    W= 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= 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: Ek=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.

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:    Wd= F x (v / 2) t

And also remember that F = m x a so:

Step 4:    Wd= ma x (v / 2) t

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

Step 5:        Wd = (v / t) x (v / 2) t

The ts cancel so:

Step 6:    Wd= ½mv2

Based on this, of course, E= 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= ½mvby 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.

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Filed under Bar Model, Dual Coding, Education, Physics

## Kinetic Energy Using The Singapore Bar Model

I think the Singapore Bar Model is a neat bit of pedagogy that has great potential in Science education.

Essentially, the Singapore Bar Model uses pictorial representations (often in the form of a bar or line) to help students bridge the gap between concrete and abstract reasoning. I wrote about one possible application here.

A recent discussion on Twitter started me thinking about if it could be applied to kinetic energy.

For example, how would you explain what happens to the kinetic energy of an object if its velocity is halved?

Many students assume that the KE would halve as well, instead of reducing to a quarter of its original value.

How can we help students grasp this slippery concept without using algebra? Algebra would work fine with your higher sets, of course, but not necessarily for other groups. This gives a clear visual representation of the fact that the KE quarters when the velocity halves. In other words, 0.5 x 0.5 = 0.25.

(Note that I have purposefully used decimals as we know that many students struggle with fractions(!))

Many students found the following question on an AQA paper extremely challenging: The correct answer is that the power output drops to one eighth of its original value.

Could the Singapore Bar Model helps students to see why this is the case?

I think it could: 