Chapter 12: Problem 2
A cube 5.0 cm on each side is made of a metal alloy. After you drill a cylindrical hole 2.0 cm in diameter all the way through and perpendicular to one face, you find that the cube weighs 6.30 N. (a) What is the density of this metal? (b) What did the cube weigh before you drilled the hole in it?
Short Answer
Expert verified
The density is approximately 5.88 g/cm³ and the initial weight was about 7.21 N.
Step by step solution
01
Calculate the Volume of the Original Cube
The volume of a cube is calculated by the formula \( V_{cube} = s^3 \) where \( s \) is the side length. Here, \( s = 5.0 \) cm. Therefore, the volume of the cube is \( V_{cube} = (5.0)^3 = 125 \text{ cm}^3 \).
02
Calculate the Volume of the Cylindrical Hole
The volume of a cylinder is calculated by the formula \( V_{cylinder} = \pi r^2 h \). The radius \( r \) is half the diameter, so \( r = 1.0 \) cm. The height \( h \) is the same as the side of the cube, which is 5.0 cm. Thus, the volume of the cylindrical hole is \( V_{cylinder} = \pi (1.0)^2 (5.0) = 5\pi \text{ cm}^3 \approx 15.71 \text{ cm}^3 \).
03
Calculate the Volume of the Remaining Cube
Subtract the volume of the cylindrical hole from the volume of the original cube to find the volume of the remaining metal. \( V_{remaining} = 125 - 5\pi \text{ cm}^3 \approx 125 - 15.71 \approx 109.29 \text{ cm}^3 \).
04
Determine the Density of the Metal Alloy
Density \( \rho \) is defined as mass per unit volume \( \rho = \frac{m}{V} \). The weight of the cube after drilling is given as 6.30 N. Assuming the density of alloy \( \rho = \frac{6.30}{9.81} \approx 0.642 \text{ kg} \), using weight \( W = mg \), gives the mass \( m \approx 0.642 \text{ kg} \). Therefore, the density \( \rho = \frac{0.642}{0.10929} \approx 5.88 \text{ g/cm}^3 \).
05
Calculate the Initial Weight of the Cube
To find the initial weight, calculate the initial mass: \( m_{initial} = 125 \text{ cm}^3 \times 5.88 \text{ g/cm}^3 \approx 735 \text{ g} \). Convert this mass to weight: \( W_{initial} = 0.735 \text{ kg} \times 9.81 \text{ m/s}^2 \approx 7.21 \text{ N} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Volume calculation
Let's dive into the concept of volume calculation, crucial for understanding how much space an object occupies. In geometry, the volume of a cube and a cylinder is typically calculated using simple formulas.
The volume of a cube is given by the formula:
The next step involves calculating the volume of a cylindrical hole drilled through the cube. The cylinder's volume is calculated with another well-known formula:
The volume of a cube is given by the formula:
- \( V_{cube} = s^3 \)
- \( V_{cube} = (5.0)^3 = 125 \text{ cm}^3 \)
The next step involves calculating the volume of a cylindrical hole drilled through the cube. The cylinder's volume is calculated with another well-known formula:
- \( V_{cylinder} = \pi r^2 h \)
- \( V_{cylinder} = \pi (1.0)^2 (5.0) = 5\pi \text{ cm}^3 \)
- Approximately \( 15.71 \text{ cm}^3 \)
- \( V_{remaining} = 125 - 5\pi \approx 109.29 \text{ cm}^3 \)
Cylindrical hole
When a cylindrical hole is drilled into a solid, it removes a specific volume of material, which can directly be calculated if the dimensions of the cylinder are known.
The key here is understanding the cylinder's dimensions:
The key here is understanding the cylinder's dimensions:
- Radius \( r \): It is half of the cylinder's diameter. Since the hole's diameter is 2.0 cm, we have \( r = 1.0 \text{ cm} \).
- Height \( h \): Since the cylindrical hole is drilled through the cube, the height of the cylinder equals the cube's side length, which is 5.0 cm.
- \( V_{cylinder} = \pi r^2 h \)
- \( V_{cylinder} = \pi (1.0)^2 (5.0) = 5\pi \text{ cm}^3 \)
- This results in approximately \( 15.71 \text{ cm}^3 \)
Weight calculation
Weight calculation is fundamental to understanding the density of a material, which is essentially the mass per unit volume.
After drilling, the cube weighed 6.30 N. Using the relation between weight and mass:
After drilling, the cube weighed 6.30 N. Using the relation between weight and mass:
- \( W = mg \)
- \( m = \frac{W}{g} = \frac{6.30}{9.81} \approx 0.642 \text{ kg} \)
- \( \rho = \frac{m}{V} = \frac{0.642}{0.10929} \approx 5.88 \text{ g/cm}^3 \)
- \( m_{initial} = 125 \text{ cm}^3 \times 5.88 \text{ g/cm}^3 \approx 735 \text{ g} \)
- \( W_{initial} = 0.735 \text{ kg} \times 9.81 \text{ m/s}^2 \approx 7.21 \text{ N} \)