# Summer Assignment Problem Help

## Homework Statement

Density + Uncertainty of block?
The mass of a rectangular block is measured to be 2.2 kg with an uncertainty of 0.2 kg. The sides are measured to be 60 +/- 3mm, 50 +/- 1 mm, and 40 +/- 2mm. Find the density of the block in kilograms per cubic meter, giving the uncertainty in the result.

D=m/V

## The Attempt at a Solution

Mass = 2.2 +/- 0.2kg
Volume = 60mm x 50mm x 40mm = 120000mm3

then... 1/50 = .02, 3/60 = .05, 2/40 = .05 ---> sum = .12

120000mm3 x .12 = 14400

D=m/V
=2.2/120000 (really not supposed to use this until i get proper conversion)
= ???

Thanks

Finding errors of things takes a little practice, as I'm sure you'll find out soon enough! As you correctly point out, Density = mass / volume. If you're multiplying or dividing, you need to use relative errors (e.g. percentages), not the absolute errors you have provided. Finding the relative error is easy though; all you need to do is take the absolute error ($dx$), and divide it by the measurement ($x$). So, for mass, the relative error is $\frac{dx}{x} =\frac{0.2}{2.2} = 0.091$. I'll let you work out the rel. errors for the dimensions.
So, if you make the Density = Mass / Vol calculation, that'll obviously give you the density. To find the error on this density, you need to add your relative errors 'in quadrature'. All this means is that you square all of your relative errors, add them together, and take the square root (i.e. $\sqrt{(\frac{dx}{x})^2 +(\frac{dy}{y})^2 + (\frac{dz}{z})^2 +... }$) This will give you the relative error of your density. If you need the absolute error of the density, you must multiply the relative error by the value you got for the density.