Question

A cube of side 0.2 m rests on the floor, as shown. Given that the cube has a mass of 50 kg, the pressure exerted by the cube on the floor is _______. (Take g = 10 $Nk{g}^{-1}$)

Answer 1

We know that F = ma
So here, the force (F) applied by the cube on the floor is:
F = mg
F = 50 $\times$ 10
F = 500 N

Now area (A) of the surface of the cube that is in contact with the floor is:
A = 0.2 $\times$ 0.2
A = 0.04 $m^2$

Since, Pressure, P = $\frac{F}{A}$

P = $\frac{500}{0.04}$

P = 12500 $N{m}^{-2}$