Question

A book rest on a table which it's face having a sides 30cm by 25cm . if it exerts apressure of 200pa then determine the mass of the block.​

Answer 1

Approximately
`1.5\; \rm kg`. (Assuming that this table is level, and that the gravitational field strength is
`g = 9.8\; \rm N \cdot kg^(-1)`.

Step-by-step explanation:

Convert the dimensions of this book to standard units:

`\displaystyle 30\; \rm cm = 30\; \rm cm * (1\; \rm m)/(100\; \rm cm) = 0.30\; \rm m`.

`\displaystyle 25\; \rm cm = 25\; \rm cm * (1\; \rm m)/(100\; \rm cm) = 0.25\; \rm m`.

Calculate the surface area of this book:

`0.30\; \rm m * 0.25\; \rm m = 0.075\; \rm m^(2)`.

Pressure is the ratio between normal force and the area over which this force is applied.

`\displaystyle \text{Pressure} = \frac{\text{normal Force}}{\text{contact Area}}`.

Equivalently:

`(\text{normal Force}) = \text{Pressure} \cdot (\text{contact Area})`.

In this question,
`\text{Pressure} = 200\; \rm Pa = 200\; \rm N \cdot m^(-2)`.

It was found that
`(\text{contact Area}) = 0.075\; \rm m^(2)` (assuming that the entire side of this book is in contact with the table.

Hence:

`\begin{aligned}& (\text{normal Force}) \\ &= \text{Pressure} \cdot (\text{contact Area}) \\ &= 200\; \rm N \cdot m^(-2) * 0.075\; \rm m^(2) \\ &= 15\; \rm N \end{aligned}`.

If that the table is level, this normal force would be equal to the weight of this book:

`\text{weight} = 15\; \rm N`.

Assuming that the gravitational field strength is
`g = 9.8\; \rm N \cdot kg^(-1)`. The mass of this book would be:

`\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= (15\; \rm N)/(9.8\; \rm N \cdot kg^(-1))\approx 1.5\; \rm kg\end{aligned}`.