Chapter 6, Problem 6.34QP

(a)

Interpretation Introduction

Interpretation: The gold block represented in figure $\text{P}6.34$ has a mass of $38.6\text{g}$. The pressure exerted by the block when it is on a square face and a rectangular face is to be calculated.

Concept introduction: The ratio of force per unit area of an object is defined as pressure. The formula to calculate pressure is,

$P=\frac{F}{A}$

The formula to calculate force is,

$F=ma$

To determine: The pressure exerted by the gold block when it is on a square face.

Answer to Problem 6.34QP

Solution

The pressure exerted by the gold block when it is on a square face is $\underset{\_}{3.8\times 10^{3}Pa}$.

Explanation of Solution

Explanation

Given

The given mass of block is $38.6\text{g}$.

The given side of the square is $1.00\text{cm}$.

Acceleration caused by gravitational force is $9.8\text{m}/{\text{s}}^2$.

Conversion of $1\text{g}$ into $\text{kg}$ is done as,

$1\text{g}=\frac{1}{1000}\text{kg}$

Therefore, conversion of $38.6\text{g}$ into $\text{kg}$ is,

$\begin{array}{c}38.6\text{g}=\frac{1\times 38.6}{1000}\text{kg}\\ \text{=0}\text{.0386}\text{kg}\end{array}$

The formula to calculate force is,

$F=ma$

Where,

• $F$ represents the force.
• $m$ represents the mass.
• $a$ represents the acceleration.

Substitute the value of mass and acceleration in above formula.

$\begin{array}{c}F=0.0386\text{kg}\times 9.8\text{m}/{\text{s}}^2\\ F=0.37828\text{kg}\cdot \text{m}/{\text{s}}^2\end{array}$

Conversion of $1\text{cm}$ into $\text{m}$ is done as,

$1\text{cm}=\frac{1}{100}\text{m}$

Therefore, conversion of $1.00\text{cm}$ into $\text{m}$ is,

$\begin{array}{c}1.00\text{cm}=\frac{1\times 1.00}{100}\text{m}\\ \text{1}\text{.00}\text{cm=0}\text{.01}\text{m}\end{array}$

The formula to calculate area of square is,

$A={\text{side}}^2$

Substitute the value of side in the above formula.

$\begin{array}{c}A={\left(0.01\text{m}\right)}^2\\ A=1\times {10}^{-4}{\text{m}}^2\end{array}$

The formula to calculate pressure is,

$P=\frac{F}{A}$

Where,

• $P$ represents the pressure.
• $F$ represents the force.
• $A$ represents the surface area.

Substitute the value of force and area in above formula.

$\begin{array}{l}P=\frac{0.37828\text{kg}\cdot \text{m}/{\text{s}}^2}{1\times {10}^{-4}{\text{m}}^2}\\ P=3.8\times {10}^3\text{kg}/\text{m}\cdot {\text{s}}^2\end{array}$

Conversion of $1\text{kg}/\text{m}\cdot {\text{s}}^2$ into $\text{Pa}$ is done as,

$1\text{kg}/\text{m}\cdot {\text{s}}^2=1\text{Pa}$

Therefore, conversion of $3.8\times {10}^3\text{kg}/\text{m}\cdot {\text{s}}^2$ into $\text{Pa}$ is,

$3.8\times {10}^3\text{kg}/\text{m}\cdot {\text{s}}^2=3.8\times {10}^3\text{Pa}$

Therefore, the value of pressure exerted by the gold block when it is on square is $\underset{\_}{3.8\times 10^{3}Pa}$

(b)

Interpretation Introduction

To determine: The pressure exerted by the gold block when it is on a rectangular face.

Answer to Problem 6.34QP

Solution

The pressure exerted by the gold block when it is on a rectangular face is $\underset{\_}{189.14Pa}$.

Explanation of Solution

Explanation

Given

The given mass of block is $38.6\text{g}$.

The given side of the rectangular is,

$\begin{array}{c}\text{Length}=2.00\text{cm}\\ \text{Breadth}=1.00\text{cm}\\ \text{Height}=1.00\text{cm}\end{array}$

Acceleration caused by gravitational force is $9.8\text{m}/{\text{s}}^2$.

Conversion of $1\text{g}$ into $\text{kg}$ is done as,

$1\text{g}=\frac{1}{1000}\text{kg}$

Therefore, conversion of $38.6\text{g}$ into $\text{kg}$ is,

$\begin{array}{c}38.6\text{g}=\frac{1\times 38.6}{1000}\text{kg}\\ =\text{0}\text{.0386}\text{kg}\end{array}$

The formula to calculate force is,

$F=ma$

Where,

• $F$ represents the force.
• $m$ represents the mass.
• $a$ represents the acceleration.

Substitute the value of mass and acceleration in above formula.

$\begin{array}{c}F=0.0386\text{kg}\times 9.8\text{m}/{\text{s}}^2\\ F=0.37828\text{kg}\cdot \text{m}/{\text{s}}^2\end{array}$

Conversion of $1\text{cm}$ into $\text{m}$ is done as,

$1\text{cm}=\frac{1}{100}\text{m}$

Therefore, conversion of $1.00\text{cm}$ and $2.00\text{cm}$ into $\text{m}$ is,

$\begin{array}{c}1.00\text{cm}=\frac{1\times 1.00}{100}\text{m}\\ \text{1}\text{.00}\text{cm}=\text{0}\text{.01}\text{m}\\ \text{2}\text{.00}\text{cm}=\text{0}\text{.02}\text{m}\end{array}$

The formula to calculate area of rectangular face is,

$A=\text{Length}\times \text{Breadth}$

Substitute the value of side in the above formula.

$\begin{array}{c}A=\left(0.01\text{m}\right)\times \left(0.01\text{m}\right)\\ A=2.00\times {10}^{-4}{\text{m}}^2\end{array}$

The formula to calculate pressure is,

$P=\frac{F}{A}$

Where,

• $P$ represents the pressure.
• $F$ represents the force.
• $A$ represents the surface area.

Substitute the value of force and area in above formula.

$\begin{array}{l}P=\frac{0.37828\text{kg}\cdot \text{m}/{\text{s}}^2}{2.00\times {10}^{-4}{\text{m}}^2}\\ P=1891.4\text{kg}/\text{m}\cdot {\text{s}}^2\end{array}$

Conversion of $1\text{kg}/\text{m}\cdot {\text{s}}^2$ into $\text{Pa}$ is done as,

$1\text{kg}/\text{m}\cdot {\text{s}}^2=1\text{Pa}$

Therefore, conversion of $1891.4\text{kg}/\text{m}\cdot {\text{s}}^2$ into $\text{Pa}$ is,

$1891.4\text{kg}/\text{m}\cdot {\text{s}}^2=189.14\text{Pa}$

Therefore, the value of pressure exerted by the rectangular face of block is $\underset{\_}{189.14Pa}$

Conclusion

The pressure exerted by the gold block when it is on a square face is $\underset{\_}{3.8\times 10^{3}Pa}$.

The pressure exerted by the gold block when it is on a rectangular face is $\underset{\_}{189.14Pa}$