Question

2. A man weighing 80 kg and wearing a flat-soled pair of shoes of area 300 cm square each is standing on a
horizontal floor. What is the pressure applied by him on the floor? What would be the pressure applied
by a women of the same weight wearing a high heeled pair of shoes of area 70 cm² each?

Answer 1

Answer:

The pressure exerted by man is $2.67\times 10^{4}Nm^{-2}$ and by woman is $1.14 \times 10^{5}Nm^{-2}$.

Step-by-step explanation:

Given the area of the shoes of a man,  $A= 300 cm^{2} =300 \times 10^{-4} m^{2}=3 \times 10^{-2} m^{2}$

Mass of the man, m = 80kg

The force applied by the man is equal to the weight of the man.

Weight is equal to the product of mass, m and acceleration due to gravity, g. i.e.,

$F=W=mg$

Let $g=10m/s^{2}$

and hence, $F=80\times 10=800N$

Pressure is defined as the perpendicular force applied over an unit area of a surface. Written as,

$P=\dfrac{F}{A}$

Substituting the values, the pressure applied by the man on the floor is

$P=\dfrac{800}{3 \times 10^{-2} }$

   $=\dfrac{8}{3 } \times 10^{4}=2.67\times 10^{4}Nm^{-2}$

Given the same weight of a woman, i.e., m = 80kg

Area of the shoes,  $A=70 cm^{2} =70 \times 10^{-4} m^{2}=7 \times 10^{-3} m^{2}$

Substituting the values, the pressure applied by the woman on the floor is

$P=\dfrac{800}{7 \times 10^{-3} }$

   $=\dfrac{8}{7} \times 10^{5}=1.14 \times 10^{5}Nm^{-2}$

Therefore, the pressure exerted by man is $2.67\times 10^{4}Nm^{-2}$ and by woman is $1.14 \times 10^{5}Nm^{-2}$.

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