Throughout this course we've emphasised the importance of always giving appropriate SI units next to numerical values. Although it's something that many students find difficult.

Now we'll walk you through the way the units can be handled, using a few examples, most of which you've already met. This will help to reinforce the principles, so that it's easier to work out units for yourself in the rest of the course.

Let's start by looking at the example we've already been discussing in this activity. We were told that a man pushed a car with a force of 415 newtons, through a distance of 18 metres, in a time of 36 seconds.

We are trying to find the mean power P.

We started by using the equation W = Fd to find the work done. There are two things that you absolutely must do when handling units and another thing that it is very useful to do. Let's consider the absolute essentials first.

The first thing you must do is to check that the values you are inputting are in appropriate SI units to start with.

In this case F was 415 newtons and d was 18 metres - both of these are SI units, so that's fine.

But if the distance had been quoted in kilometres or miles (or more likely in centimetres or inches if I was pushing the car!) you would need to convert these units to metres before doing the calculation. If you put in values in appropriate SI units then the answer should come out in appropriate SI units too!

It can be very tempting, especially when you're in a hurry to just write down W = 415 x 18 = 7470. But that is completely meaningless. 7470 what? You may think that this is being unduly fussy, but it is a serious business. In 1998 the Mars Climate Orbiter was accidentally destroyed in the Martian atmosphere because of a problem with units. Data was calculated by one team of scientists with one system of units (the so called 'Imperial System') and reported to another team who were expecting the data to be in a different system of units (SI - the system which is usually used in science and which we use in this course).

So you must label the units carefully, and on each line of your working. In this case, you're finding work done, which is a form of energy and so it is measured in joules. So, at the very least, you must put the units at the end of each line of your working.

But there is an even better way of handling units.

Rather than just writing joules at the end of each line of working - because you know, or think that these are the correct units of the answer - you could put the units next to each quantity that you're considering.

So you'd write W = 415 N x 18 m

Then you can actually work out the units of the answer. We can handle the numbers and the units separately to write W = 415 x 18 newtons x metres

415 x 18 is 7470 as before, and since a joule is defined to be the energy transferred when a force of one newton acts for a distance of one metre, 1 J = 1 N x 1 m so instead of the N m we can write 'J'.

We've worked out the units of the answer to be joules... and since this turns out to be the units we were expecting, this is good news. This provides a powerful way of checking the answer. If the units HAD NOT worked out to be joules, we'd have known we had done something wrong.

As another example of the way in which you can work out the units, here's the second half of the calculation in the same example. Here we were using the fact that power is the energy converted (E) divided by the time taken (t).

In this case we now have E = 7470 J and t = 36 s

So we can write P = 7470 J / 36 s

Once again handling the numbers and the units separately, 7470 divided by 36 is 210 to two significant figures and the units are joules divided by seconds which we can write as J s^-1.

But we know that a watt is a joule per second so the units have worked out to be watts and we can write P = 2.1 x 10² W. Again, this is what we expected which is good news. Suppose now that instead of calculating a value for power from known values of transferred energy and time, we had been given values for power and energy transferred, and were trying to find the time taken.

Let's suppose P = 1.5 kW and E = 9.0 x 10⁴ J.

To start with we need to convert the kilowatts to watts. 1 kW = 1 x 10³ W, so 1.5 kW is 1.5 x 10³ W.

We could rearrange the equation P = E / t to give t = E / P, you've done that before. If you can't remember how to do it I suggest that you look back to Chapter 3.

For now I want to concentrate on the units.

We can substitute the values we've been given for P and E complete with their units, so t = 9.0 x 10⁴ J / 1.5 x 10³ W

We can handle the numbers and the units separately, 9.0 x 10⁴ divided by 1.5 x 10³ gives 60, so it's 60 J / W.

But we know that 1 W is 1 J s^-1, so we can write the units in the calculation as J divided by J s^-1.

The joules cancel so we have units of 1 / s^-1.

What does 1 / s^-1 mean? Remember that 1 over anything can be written as that quantity to the power minus 1.

It follows from this that 1 / x^-1 = x, so 1 / s^-1 = s.

So the units have worked out to be seconds. Since t is time this is good news - the units have worked out as expected.

So far we have just considered cases where the units of the final answer 'drop out' quite easily from the definition of the joule and the watt, but we can take this method further by remembering that all SI units can be expressed in terms of SI base units. So far we've just introduced three base units, the metre for length, the kilogram for mass and the second for time. All the other units we've introduced so far can be expressed in terms of these.

Remember the equation for kinetic energy from Chapter 3 E_k = (1/2) mv²

Suppose you wanted to find the kinetic energy of an object of mass 10 kg and speed 2 m s^-1. So we can write E_k = (1/2) x 10 kg, putting it in brackets to make it clearer, x 2 m s^-1, all squared.

Let's start by working the number out. 2 squared is 4 so we've got (1/2) x 10 x 4 which is 20. Now the units. There's a kg from the mass term then we have to remember to square the units of v as well as the 2, so we've got kg x (m s^-1)².

The squared outside the bracket means that we have to square both the m and the s^-1, so we've got kg times m² times s^-1 squared which is s^-2.

If you look back to Chapter 3 or to your personal glossary you will see that 1 kg m² s^-2 is the same as 1 joule, so we can write E_k = 20 J.

Hey presto, the units have worked out to be joules, the SI units of energy, as expected.

Our final example concerns the rearranged form of the kinetic energy equation with v as the subject. You've met this on several occasions.

Let's suppose that you are trying to find the speed of an object of kinetic energy 50 joules and mass 4 kilograms.

These values are in SI units, so we can substitute v = √(2 x 50 J / 4 kg)

Let's consider the numbers first. 2 times 50 is 100, and 100 divided by 4 is 25 so we can write v = √(25(J / kg)).

To work the units out, remember that 1 J = 1 kg m² s^-2

So we can write v = √(25(kg m² s^-2/kg)).

The kg cancel

So we're left with v = √(25 m² s^-2)

Now the positive square root of 25 is 5 (since 5 x 5 is 25) and similarly, we know that m s^-1 all squared is m² s^-2, so the square root of m² s^-2 is m s^-1.

So v = 5 m s^-1

The units have worked out to be metres per second, which is what we might have expected since v is a speed.

But this is very different from just assuming the units to be m s^-1. We have worked these units out, in the same way as we have worked the number out to be 5. And because the units are the units that we expected for speed, we have checked that v = √(2E_k / m) is a reasonable equation to be using. This isn't a perfect way of checking the answer - it is perfectly possible to get the units right whilst getting the numerical answer wrong, but it is a good first step - if the units had not turned out to be m s^-1 we'd have known that there was something wrong.

So this method of dealing with units both reminds you that you are dealing with real scientific quantities not just numbers and it gives you a good method for checking your answer. It really is worth persevering with!