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

Answer:

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

Explanation:

Convert the dimensions of this book to standard units:

$\displaystyle 30\; \rm cm = 30\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm} = 0.30\; \rm m$.

$\displaystyle 25\; \rm cm = 25\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm} = 0.25\; \rm m$.

Calculate the surface area of this book:

$0.30\; \rm m \times 0.25\; \rm m = 0.075\; \rm m^{2}$.

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:

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

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:

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