question_answer

A cubical block of mass 2 kg with each side of 2cm is lying on the ground. What will be the pressure exerted by the block on the ground? (Take g \[=9.8\text{ }m/{{s}^{2}}\]).

Answer:

mass of the block \[(m)=2\text{ }kg\] weight of the block \[(w)=mg\] \[=2\text{ }kg\times 9.8\text{ }m/{{s}^{2}}\] \[=19.6\text{ }kg\text{ }m/{{s}^{2}}\] \[=19.6\text{ }N\] Side of cube \[(a)=2\text{ }cm=\frac{2}{100}m=0.02\,m\] Area of cube                 \[={{a}^{2}}=0.02m\times 0.02\text{ }m=4\times {{10}^{4}}{{m}^{2}}\] \[\therefore \]\[P=\frac{F}{A}=\frac{weight}{Area}=\frac{19.6\,N}{4\times {{10}^{-4}}{{m}^{2}}}\] \[=4.9\times {{10}^{4}}N/{{m}^{2}}=4.9\times {{10}^{4}}Pa\]