A block of 10x20x25 cm and weighing 5 kg lies on the table with its larger edge. What kind of pressure does he create?

Let a block having the shape of a rectangular parallelepiped weighing m = 5 kg has dimensions: 10x20x25 cm.It is known from the problem statement that the block lies on the table with its larger facet, then the large facet of the bar will have an area:

S = 20 cm ∙ 25 cm = 500 cm ^ 2;

S = 0.05 m ^ 2.

From the side of the bar, a force F acts on the table, equal to its weight:

F = m ∙ g, where coefficient g = 9.8 H / kg.

To determine what pressure he creates on the table, we find the ratio:

p = F: S or p = m ∙ g: S.

Substitute the values ​​of physical quantities in the calculation formula:

p = 5 kg ∙ 9.8 H / kg: (0.05 m ^ 2);

p = 980 Pa.

Answer: the bar creates a pressure of 980 Pa on the table.

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