Question

A cubical block of wood of length $5cm$ is kept on a table top. The density of the block is $800kg/m^2$ and $g=9.8m/s^2$. Find the pressure exerted by the wooden block on the table top.

Answer 1

First we find mass $\left(m\right)$.
$density\left(d\right)=\frac{mass}{volume}$
$\therefore 800=\frac{m}{125c{m}^3}=\frac{m}{125\times {10}^{-6}{m}^3}$
$\Rightarrow m=(800\times 125\times {10}^{-6})kg=0.1kg$
Now, pressure $\left(P\right)=\frac{F}{A}=\frac{mg}{A}=\frac{(0.1\times 9.8)N}{5cm\times 5cm}$
$=\frac{(0.1\times 9.8)N}{0.05m\times 0.05m}=292N/m^2=292Pa$.