A person of mass 55 kg stands momentarily on one heel while walking. The heel has a contact area with the floor of 1.2 cm^2 (1.2×10^-4 m^2). Using g = 9.81 m s^-2, what pressure does the heel exert on the floor?

AS Level Physics 9702: Pressure Exerted by a Shoe Heel

Setting up the problem

Pressure is force per unit area, $p = F/A$ , where $F$ is the force acting perpendicular to the surface. Here the force is the person’s weight, not their mass, so it must first be calculated: $F = mg = 55 \times 9.81 = 539.55\text{ N}$

The contact area is already given in m²: $A = 1.2\times10^{-4}\text{ m}^2$ .

Calculating the pressure

$p = \frac{F}{A} = \frac{539.55}{1.2\times10^{-4}}$ $p = 4.49625\times10^{6}\text{ Pa} \approx 4.50\times10^{6}\text{ Pa (3 s.f.)}$

Why the other options are wrong

A ( $4.58\times10^{5}\text{ Pa}$ ): uses the mass directly in place of the weight, i.e. calculates $m/A = 55/(1.2\times10^{-4})$ instead of $mg/A$ .
C ( $4.58\times10^{6}\text{ Pa}$ ): uses $g = 10\text{ m s}^{-2}$ instead of the given $g = 9.81\text{ m s}^{-2}$ : $(55\times10)/(1.2\times10^{-4}) = 4.58\times10^{6}\text{ Pa}$ .
D ( $4.50\times10^{2}\text{ Pa}$ ): forgets to convert the area from cm² to m², dividing by $1.2$ instead of $1.2\times10^{-4}$ .

Final answer

Pressure exerted by the heel $= \boxed{4.50\times10^{6}}\text{ Pa}$ , option B.