A $1.00 m^3$ volume of iron has a mass of $7.86 \cdot 10^3 kg$, find the length of one side with a mass of $200g$

Question

So this is a relatively simple physics problem, but I'm sort of stumped and need some help finishing this off.

The full problem reads:

Iron has a property such that a $1.00 m^3$ volume has a mass of $7.86\cdot10^3 kg$ (density equals $7.86 \cdot 10^3 kg/m^3$). You want to manufacturer iron into cubes and spheres. Find the length of the side of a cube of iron that has a mass of $200.0g$.

So the first thing we do is try to solve for the volume of a cube with a mass of $200.0g$.

$$ .200kg \frac{1 m^3}{7.86\cdot10^3kg} = 2.94 \cdot 10^{-5} m^3$$

At this point I get lost. When I look at the book answer they just immediately conclude that

$$ 2.94\cdot 10^{-5}m^3$$ leads directly to $$2.94\cdot10^2 m$$

But I'm not sure how they eliminated the other units from the volume.

Can anyone help me? Especially if you could explain it using dimensional analysis because I would really like to get better at it.

Thank you!

Answer 1

Firstly you shouldn't use $x$ for your scientific notation --- use $\times$ (\times). Also a good convention with units is to leave spaces between the number and the unit (and other units)... for example

$20\,\text{m s}^{-1}$ is better than $20\text{ms}^{-1}$.

So anyway we have a density of $\rho=7.86\times 10^3$ kg m$^{-3}$.

Note we define density as

$$\rho=\frac{m}{v},$$

and the volume of a cube of side $s$ is $s^3$ (the sphere is a red herring). Therefore our equation reads

$$\rho=\frac{m}{s^3}.$$

Recall we want to find $s$ --- the length of a side. A lot of students at this point substitute all the numbers at this point... keep all that messing until the end!

Now a way to solve this for $s$ is

$$\begin{align}\rho&=\frac{m}{s^3} \\ \Rightarrow \rho s^3&=m \\ \Rightarrow s^3&=\frac{m}{\rho} \\ \Rightarrow s&=\sqrt[3]{\frac{m}{\rho}}\end{align}$$ (there are two complex solutions to the abstract problem but only one physical one --- positive roots are unique and $s$ is certainly positive!)

Your biggest problem now is to get your units in order... using the units there note we have

$$\rho=\sqrt[3]{\frac{200\text{ g}}{7.86\times10^3\text{ kg m}^{-3}}}.$$

Note we don't have $\displaystyle\frac{kg}{kg}=1$, leaving $\frac{1}{\text{m}^{-3}}=\text{m}^3$, which, once we take the cube root the correct units m.

One way of getting around this is realising that for many equations, using SI units renders this unit analysis unnecessary --- convert the 200 g to kilograms either in your head or via

$$\begin{align}1000\text{ g}&=\text{kg} \\ \Rightarrow \text{g}&=\frac{\text{kg}}{1000} \\ \Rightarrow 200\text{ g}&=200\,\frac{\text{kg}}{1000}=0.2\text{ kg}\end{align},$$

so we have

$$s=\sqrt[3]{\frac{0.2}{7.86\times10^3}}\approx 0.02941\text{ m}\approx 3\text{ cm}.$$

Answer 2

The volume you compute is not correct. It should be $\frac {0.200 \text{kg}}{7860 \text{kg/m}^3}\approx 0.0000254 \text {m}^3 =2.54\cdot 10^{-5} \text {m}^3$ To get the side of a cube with that volume, we take the cube root and in fact $\sqrt[3]{2.54\cdot 10^{-5} \text {m}^3} \approx 0.0294 \text {m}$

Answer 3

Area varies as the square of length of cube or radius of a sphere and, Volume varies as the cube of length of a cube or radius of a sphere.

If s and b are subscripts for small and big, the proportion is:

$$ \left( \frac{L_s}{L_b} \right)^3= \frac {M_s}{M_b} $$

or, $ L_s = L_b \left(\frac {M_s}{M_b} \right)^{\frac13} = 1.0\, (\frac{.5}{7.86\,10^3})^{1/3}=0.029413 $ meters.