Problem 6 - College Physics

Question

A cube 5.0 \(\mathrm{cm}\) on each side is made of a metal alloy. After you drill a cylindrical hole 2.0 \(\mathrm{cm}\) in diameter all the way through and perpendicular to one face, you find that the cube weighs 7.50 \(\mathrm{N}\) (a) What is the density of this metal? (b) What did the cube weigh before you drilled the hole in it?

Step-by-Step Solution

Answer

Density: 7.00 g/cm³; Original weight: 8.58 N.

Step 1: Calculate the Volume of the Cube

The cube has a side length of 5.0 cm. To find the volume of the cube, use the formula for the volume of a cube:\[ V_{\text{cube}} = s^3 \]where \( s \) is the side length of the cube.Substitute \( s = 5.0 \) cm into the formula:\[ V_{\text{cube}} = (5.0)^3 = 125.0 \, \text{cm}^3 \]

Step 2: Calculate the Volume of the Cylinder Removed

The cylindrical hole drilled through the cube has a diameter of 2.0 cm, which means it has a radius \( r = 1.0 \) cm, as radius is half the diameter. The height of the cylinder is equal to the side of the cube, 5.0 cm. Use the formula for the volume of a cylinder:\[ V_{\text{cylinder}} = \pi r^2 h \]Substitute \( r = 1.0 \) cm and \( h = 5.0 \) cm:\[ V_{\text{cylinder}} = \pi (1.0)^2 (5.0) = 5\pi \, \text{cm}^3 \]

Step 3: Calculate the Volume of the Metal After the Hole is Drilled

To find the volume of the metal after drilling, subtract the volume of the cylindrical hole from the volume of the cube:\[ V_{\text{metal}} = V_{\text{cube}} - V_{\text{cylinder}} \]Substitute the known values:\[ V_{\text{metal}} = 125.0 - 5\pi \approx 109.28 \, \text{cm}^3 \]

Step 4: Calculate the Density of the Metal

Density \( \rho \) is defined as mass divided by volume. First, find the mass using weight and the relation \( F = mg \), where \( F \) is the weight and \( g \) is the acceleration due to gravity (9.8 m/s²):\[ m = \frac{F}{g} = \frac{7.50}{9.8} \, \text{kg} \approx 0.7653 \, \text{kg} \]Now, convert the mass to grams (1 kg = 1000 g):\[ m = 0.7653 \, \text{kg} = 765.3 \, \text{g} \]Density formula:\[ \rho = \frac{\text{mass}}{\text{volume}} \]Substitute mass and the volume of metal:\[ \rho = \frac{765.3}{109.28} \approx 7.00 \, \text{g/cm}^3 \]

Step 5: Calculate the Mass of the Original Cube

Using the density found in Step 4, we can calculate the mass of the original cube (before the hole was drilled). Use the volume of the entire cube:\[ m_{\text{original}} = \rho \times V_{\text{cube}} \]\[ m_{\text{original}} = 7.00 \times 125.0 = 875.0 \, \text{g} \]

Step 6: Convert the Mass Back to Weight of the Original Cube

To find the original weight in newtons, use the formula \( F = mg \):\[ F_{\text{original}} = 875.0 \, \text{g} \times \frac{1 \, \text{kg}}{1000 \, \text{g}} \times 9.8 \, \text{m/s}^2 \approx 8.575 \, \text{N} \]

Step 7: Conclusion

The density of the metal is approximately 7.00 g/cm³, and the original weight of the cube was approximately 8.58 N.

Key Concepts

Volume Calculation

The calculation of volume is a fundamental concept in geometry and physics, especially when dealing with objects of regular shape. Volume measures how much space an object occupies. To calculate volume, different formulas apply to different shapes. For example:
- **Cube**: The volume of a cube is calculated by taking the cube of its side length. Mathematically, this is given by \( V_{ ext{cube}} = s^3 \), where \( s \) is the side length. For our cube with a side length of 5.0 cm, the volume is \( 125.0 \text{ cm}^3 \).
- **Cylinder**: The volume of a cylinder is given by \( V_{ ext{cylinder}} = \pi r^2 h \), where \( r \) is the radius of the base and \( h \) is the height. A drilled cylindrical hole with a radius of 1.0 cm and height of 5.0 cm results in a volume of \( 5\pi \text{ cm}^3 \). Understanding these relationships helps in finding how much material makes up an object or how much it can contain.

Mass and Weight

Mass and weight are often confused but are distinct concepts in physics. **Mass** is a measure of how much matter an object contains and is usually measured in grams or kilograms. It does not change based on location.
**Weight**, on the other hand, is the force exerted by gravity on an object. It is calculated using the formula \( F = mg \), where \( F \) is the force (or weight) in newtons, \( m \) is the mass in kilograms, and \( g \) is the acceleration due to gravity, typically \( 9.8 \text{ m/s}^2 \) on Earth. After drilling the hole, the cube's mass was determined using its weight and rearranging the formula: \( m = \frac{F}{g} \). Converting between mass and weight is crucial for understanding how they interact under gravitational forces.

Cylinder Volume

A cylinder is a three-dimensional shape with two parallel circular bases and a height. Calculating the volume of a cylinder is key in many scientific and engineering problems. The formula \( V_{ ext{cylinder}} = \pi r^2 h \) captures this, involving:

• **\( \pi \)**: A constant approximately equal to 3.14159.
• **\( r \)**: The radius of the cylinder's base.
• **\( h \)**: The height of the cylinder.

In our exercise, the cylindrical hole has a radius of 1.0 cm and a height of 5.0 cm, resulting in a volume of \( 5\pi \text{ cm}^3 \). This measurement helps in determining how much of the original cube’s material is removed and provides insight into the configuration of objects.

Metal Density

Density is a measure of how compact the mass in a substance is. It is commonly expressed in units such as grams per cubic centimeter (g/cm³). For metals, knowing the density is important for their identification and use in construction and manufacturing.
The formula for calculating density is \( \rho = \frac{\text{mass}}{\text{volume}} \). In the exercise, the metal density of the modified cube was calculated by substituting the cube's mass (765.3 g) and volume after the hole was drilled (109.28 cm³). This led to a density of approximately 7.00 g/cm³.
Understanding metal density aids in tasks like calculating weight for shipping, ensuring material strength, and selecting the right metal for applications based on its density.