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 _ __ _

0.2m

(Take g = 10 N kg 1

)

A. 25 N m 2 B. 250 N m

2

C. 1250 N m

2 D. 12500 N m-2

Answer 1

we know F= ma
here force applied by the cube on the floor (F)=50×10=500N

now are of the cube in contact with the surface
A=0.2×0.2=0.04m2

Also pressure(P)=F/A
so here P =500N/0.04m2
=12500pascal


Therefore, option D is the answer.

hope it helps.....

Answer 2

Pressure exerted by the cube on the floor is (D) 12500 N/m².

Given:

Side of cube = 0.2 m

Mass of cube = 50 kg

To find:

Pressure exerted by cube on the floor.

Solution:

We know pressure is the force applied on an object per unit area.

Mathematically, $Pressure (P) = \frac{Force (F)}{Area (A)}$

Step 1

Force applied by cube on the floor will be nothing but the weight of the cube.

That is,

$F = mg$

Substituting the given values, we get

$F = 50 (10)\\F = 500 N$

Step 2

Area of the cube

Here, the total area will not be considered, as due to the weight of the cube, the pressure is applied on the ground and the area that is in contact with the ground will only be considered.

Hence,

Area = $(side)^{2}$

side = 0.2 m

Therefore,

$Area = (0.2)^{2} \\Area = 0.04 m^{2}$

Therefore, the pressure will be,

$P = \frac{500}{0.04} \\P = 12500 N/m^{2}\\P = 12.5 K N/m^{2}$

Final answer:

Hence, the pressure exerted by the cube on the floor is $12500 N/m^{2}$(D).