Problem 65 - University Physics with Modern Physics

Question

Iron has a property such that a \(1.00-\mathrm{m}^{3}\) volume has a mass of \(7.86 \times 10^{3} \mathrm{kg}\) (density equals \(7.86 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3} ) .\) You want to manufacture iron into cubes and spheres. Find (a) the length of the side of a cube of iron that has a mass of 200.0 \(\mathrm{g}\) and (b) the radius of a solid sphere of iron that has a mass of 200.0 \(\mathrm{g}\) .

Step-by-Step Solution

Answer

a) The side length of the cube is approximately 0.02993 m. b) The radius of the sphere is approximately 0.01839 m.

Step 1: Understanding the Problem

We need to find the dimensions of an iron cube and sphere that each have a mass of 200 grams. Given that the density of iron is \(7.86 \times 10^3 \text{ kg/m}^3\), we'll find the volume first and then use that to find the side length of the cube and the radius of the sphere.

Step 2: Convert Mass to Kilograms

Convert the given mass from grams to kilograms because the density is provided in kg/m^3. \[ 200 \text{ g} = 0.200 \text{ kg} \]

Step 3: Calculate the Volume of Iron

Use the formula for density to find the volume. Density (\( \rho \)) is mass over volume, \( \rho = \frac{m}{V} \), thus the volume \( V = \frac{m}{\rho} \).\[ V = \frac{0.200 \text{ kg}}{7.86 \times 10^3 \text{ kg/m}^3} = 2.5442 \times 10^{-5} \text{ m}^3 \]

Step 4: Find the Side Length of the Cube

For a cube, the volume is \( s^3 = V \), where \( s \) is the side length. Solve for \( s \).\[ s = \sqrt[3]{V} = \sqrt[3]{2.5442 \times 10^{-5} \text{ m}^3} \approx 0.02993 \text{ m} \]

Step 5: Find the Radius of the Sphere

For a sphere, the volume is related to the radius \( r \) by the equation \( V = \frac{4}{3} \pi r^3 \). Solve for \( r \). \[ r^3 = \frac{3V}{4\pi} = \frac{3(2.5442 \times 10^{-5})}{4\pi} \]\[ r = \sqrt[3]{\frac{3(2.5442 \times 10^{-5})}{4\pi}} \approx 0.01839 \text{ m} \]

Key Concepts

Mass and Volume Relationship

Understanding the relationship between mass and volume is crucial in physics, especially when dealing with materials like iron. Mass refers to the amount of matter in an object, measured in kilograms (kg), while volume indicates the space that an object occupies, measured in cubic meters (m³). The density of a material tells us how much mass is contained within a specific volume. For example, iron has a density of \(7.86 \times 10^3 \text{ kg/m}^3\), meaning each cubic meter of iron weighs \(7.86 \times 10^3\) kg.
This relationship is expressed by the formula for density: \( \rho = \frac{m}{V} \), where \( \rho \) is density, \( m \) is mass, and \( V \) is volume. Rearranging this formula helps us find the volume if the mass and density are known: \( V = \frac{m}{\rho} \).
This principle allows us to determine dimensions of objects, such as the side of a cube or the radius of a sphere, based on their mass and material density.

Geometric Shapes in Physics

Geometric shapes, such as cubes and spheres, are often used in physics to simplify complex problems. These shapes have standard formulas to calculate volume, which is essential for understanding their physical characteristics.
For cubes, the volume is determined by the equation \( V = s^3 \), where \( s \) is the side length. This means if you want to find out how much space a cube takes up, you need to cube the length of one of its sides.
For spheres, the volume is given by the formula \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius. Spheres are symmetrical and such formulas are crucial for solving problems involving objects of this shape.

• Cube Formula: \( V = s^3 \)
• Sphere Formula: \( V = \frac{4}{3} \pi r^3 \)

These volume formulas are vital for calculating dimensions when dealing with materials of a known mass and density, like in the exercise with iron.

Unit Conversion in Physics

Unit conversion is a fundamental tool in physics, ensuring that different measurements align with each other. Understanding and correctly converting units can avoid confusion and make calculations more straightforward.
In the realm of density calculations, mass is often given in grams (g), but density might involve kilograms (kg) per cubic meter (\( \text{m}^3 \)). Hence, converting mass from grams to kilograms is needed. Since \( 1 \text{ kg} = 1000 \text{ g} \), we convert by dividing by 1000: \( 200 \text{ g} = 0.200 \text{ kg} \).
Accurate unit conversion allows us to use formulas correctly and ensures that the results to problems are both precise and reliable. This accuracy is crucial when using formulas like \( \rho = \frac{m}{V} \) to determine volumes, as seen when calculating the dimensions of geometric shapes.