I am writing a lab for my high school physics class, and I calculated the change in gravitational potential energy in one trial to be $\Delta E_\text{P}=0.27\text{ J}$. The height from which the mass dropped was only measured to 2 significant figures, with an uncertainty of 1 cm. Now, in order to calculate the uncertainty in the measurement, I did the following. $$\frac{\Delta \left(\Delta E_\text{P}\right)}{\Delta E_\text{P}}=\left[\frac{\Delta\left(\Delta h\right)}{\Delta h}\right]\Longleftrightarrow\Delta \left(\Delta E_\text{P}\right)=\Delta E_\text{P}\left[\frac{\Delta\left(\Delta h\right)}{\Delta h}\right]=mg\Delta h\left[\frac{\Delta\left(\Delta h\right)}{\Delta h}\right]=mg\Delta\left(\Delta h\right)=\left(0.050\text{ kg}\right)\left(9.81\text{ m s}^{-2}\right)\left(0.01\text{ m}\right)=0.004905\text{ J}$$

In significant figures, this would be expressed as 0.005 J; however, my calculation for the change in gravitational potential energy only has 2 decimal places of precision, meaning I would further have to round the uncertainty to 0.01 J. I would be ok with this, except that 0.004905 is closer to 0 than it is to 1. Which value should I express my uncertainty as?

If I express it as 0, should I give precision to the 0? For example, 0 or 0.00 J.

Also, I know there are way better ways to express precision and significant figures, but this is what we are taught to do in class so I would like to stick to that.

  • $\begingroup$ Leading zeroes do not count for significant digits in a multiplication or division problem. $\endgroup$
    – Muphrid
    Commented Oct 8, 2016 at 19:02

2 Answers 2


First problem: you're re-rounding. Never do that. Significant figures only affect how you write a number, not what it really is. You can write a number to any given precision, but you should always start with the original, full-precision value and round that.

In this case, you first write 0.004905 with one significant figure, so you round it to 0.005. Then you want to write it with hundredths' precision. Don't start with 0.005, because that's not your number, that's just an abbreviated expression of your number. You should start with the original number, 0.004905, and round that to two decimal places, which gives you 0.00.

The second problem is that that's not two significant figures. Leading zeros are not significant. When you want to find which digits are significant, skip all the leading zeros and start where there is a nonzero digit, namely the 4 (which gets rounded up).


It may help if you think in terms of percentage errors?

Working from your result of $\Delta E_{\rm P}$ the change in height was $0.55$ m

So what you have is a mass of $0.0500 \pm 0.0001 $kg (I have estimated that you know the mass to an accuracy of $\pm 1$ gramme), gravitational field strength $9.81 \pm 0.01 {\rm ms^{-2}}$ and height $0.55 \pm 0.01 {\rm m}$.

Using the values that I have chosen $\Delta E_{\rm P} = 0.0500 \times 9.81 \times 0.55 = 0.269775$ J.
I have not rounded this value until the error has been calculated.

You can find the maximum percentage error by summing the percentage error in each of the quantities.

Maximum percentage error is

$$ \left (\dfrac {0.0001}{0.0500}+\dfrac {0.01}{9.81}+\dfrac {0.01}{0.55}\right ) \times 100 = (0.002+0.001+0.018) \times 100 \approx 2 \%$$

So $2 \%$ of $0.269775$ is $0.005$ so you can quote your change in potential energy as $0.270 \pm 0.005$ J.

Note that you can immediately see that to improve the accuracy of your experiment you should concentrate on measuring the height more accurately.

I hope that this shows you that it is significant figures and not decimal palces which are important.

$0.00456$ is quoted to five decimal places but only three significant figures.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.