Problem 7 - Applied Physics

Question

A block of gold \(9.00\) in. \(\times 8.00\) in. \(\times 6.00\) in. weighs \(302 \mathrm{lb}\). Find its weight density.

Step-by-Step Solution

Answer

The weight density is approximately 0.699 lb/in³.

Step 1: Understand the Problem

We are given the dimensions of a block of gold and its weight, and we need to find the weight density. Weight density is defined as weight per unit volume.

Step 2: Calculate the Volume

First, calculate the volume of the block using the formula for the volume of a rectangular prism: \( V = ext{length} imes ext{width} imes ext{height} \). Substitute the given values:\[V = 9.00 ext{ in.} imes 8.00 ext{ in.} imes 6.00 ext{ in.} = 432 ext{ cubic inches}.\]

Step 3: Calculate Weight Density

Weight density is calculated by dividing the weight by the volume of the block. The formula is:\[\text{Weight Density} = \frac{\text{Weight}}{\text{Volume}}.\]Substitute the given weight and calculated volume into the formula:\[\text{Weight Density} = \frac{302 ext{ lb}}{432 ext{ in}^3} \approx 0.699 ext{ lb/in}^3.\]

Step 4: State the Result

The weight density of the block of gold is approximately \(0.699\) lb/in³.

Key Concepts

Weight density

Weight density is an essential concept in physics, especially when dealing with problems involving materials and substances. It represents how much weight is contained in a specific volume of a substance. In other words, it tells us how many pounds or kilograms are packed into a unit of space, like a cubic inch or cubic centimeter.
To calculate weight density, you use the formula: \[\text{Weight Density} = \frac{\text{Weight}}{\text{Volume}}.\]This formula suggests that you need two pieces of information: the weight of the object and its volume.
The relationship between weight and volume in this context is similar to density, but instead of mass, you consider weight, which includes the effect of gravity.
In practical applications, knowing the weight density can help determine if a substance can support a particular load or how it will behave in different conditions. It is a crucial parameter in engineering, construction, and manufacturing.

Volume calculation

Calculating volume is a foundational skill in physics and mathematics. To find the volume of an object, you rely on formulas specific to the shape of the object. For instance, the volume of a rectangular prism, like a block of gold, can be found using the formula: \[V = \text{length} \times \text{width} \times \text{height}.\]Here, the volume is measured in cubic units, which could be cubic inches or cubic meters, depending on the units used for the dimensions.
In our exercise, to calculate the volume of the block of gold, you multiply its length, width, and height. This technique gives you a clear picture of the amount of space that the block occupies. Knowing the volume is critical, as it is directly needed to compute other physical quantities like weight density.
Remember:

• Make sure all measurements are in the same unit before multiplying.
• The result will provide a value in cubic units based on the units used for the dimensions.

Unit conversion in physics

Unit conversion is a vital skill in physics, especially when measurements are given in various units. Converting units ensures that your calculations are accurate and consistent. In our problem, while the dimensions are given in inches and weight in pounds, consistency in units is important for computing the weight density.
Why convert units?

• To ensure that calculations are made in a coherent system of units, avoiding errors.
• To compare calculated values with known values which might be in a different unit system.
• To apply formulas that require specific unit inputs.

For instance, if you ever need to convert inches to centimeters, or pounds to kilograms, using conversion factors like 1 inch equals 2.54 centimeters or 1 pound equals 0.453592 kilograms can be helpful.
Although our example didn't require a unit conversion, being prepared to convert units ensures that you can handle diverse problems, where measurements might need aligning with a specific unit system. Proper unit conversion can greatly simplify solving complex problems in physics!