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A solid concrete block weighs 150. N and is resting on the ground. Its dimensions are 0.400 m ✕ 0.200 m ✕ 0.100 m. A number of identical blocks are then stacked on top of this one. What is the smallest number of whole blocks (including the one on the ground) that can be stacked so that their weight creates a pressure of at least two atmospheres on the ground beneath the first block?

User NgLover
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1 Answer

7 votes

Answer:

27 blocks

Step-by-step explanation:

First, the expression to use here is the following:

P = F/A

Where:

P: pressure

F: Force exerted

A: Area of the block.

Now , we need to know the number of blocks needed to exert a pressure that equals at least 2 atm. To know this, we should rewrite the equation. We know that certain number of blocks, with the same weight and dimensions are putting one after one over the first block, so we can say that:

P = W/A

P = n * W1 / A

n would be the number of blocks, and W1 the weight of the block.We have all the data, and we need to calculate the area of the block which is:

A = 0.2 * 0.1 = 0.02 m²

Solving now for n:

n = P * A / W1

The pressure has to be expressed in N/m²

P = 2 atm * 1.01x10^5 N/m² atm = 2.02x10^5 N/m²

Finally, replacing all data we have:

n = 2.02x10^5 * 0.02 / 150

n = 26.93

We can round this result to 27. So the minimum number of blocks is 27.

User Erv Walter
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