Question

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

The density is approximately 5.88 g/cm³ and the initial weight was about 7.21 N.

Step-by-step solution

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 \( V_{cube} = (5.0)^3 = 125 \text{ cm}^3 \).

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 \( V_{cylinder} = \pi (1.0)^2 (5.0) = 5\pi \text{ cm}^3 \approx 15.71 \text{ cm}^3 \).

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. \( V_{remaining} = 125 - 5\pi \text{ cm}^3 \approx 125 - 15.71 \approx 109.29 \text{ cm}^3 \).

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 \( \rho = \frac{0.642}{0.10929} \approx 5.88 \text{ g/cm}^3 \).

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: \( W_{initial} = 0.735 \text{ kg} \times 9.81 \text{ m/s}^2 \approx 7.21 \text{ N} \).

Key concepts

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:

• \( V_{cube} = s^3 \)

where \( s \) is the side length of the cube. In our exercise, the cube has a side length of 5.0 cm, leading to:

• \( V_{cube} = (5.0)^3 = 125 \text{ cm}^3 \)

This means our cube occupies a total space of 125 cubic centimeters.

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:

• \( V_{cylinder} = \pi r^2 h \)

For the cylinder, \( r \), the radius, is half the diameter of 2.0 cm, which is 1.0 cm. The height \( h \) is the same as the side of the cube, 5.0 cm. Therefore, the hole has a volume of:

• \( V_{cylinder} = \pi (1.0)^2 (5.0) = 5\pi \text{ cm}^3 \)
• Approximately \( 15.71 \text{ cm}^3 \)

After determining these volumes, we can find the volume of material remaining by subtracting the cylindrical hole volume from the cube:

• \( V_{remaining} = 125 - 5\pi \approx 109.29 \text{ cm}^3 \)

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:

• 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.

Using these dimensions, we apply the volume formula for a cylinder:

• \( V_{cylinder} = \pi r^2 h \)
• \( V_{cylinder} = \pi (1.0)^2 (5.0) = 5\pi \text{ cm}^3 \)
• This results in approximately \( 15.71 \text{ cm}^3 \)

The subtraction of this volume from the original cube's volume is essential to calculate the new density because the remaining material's volume is modified by the removal of the drilled section.

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:

• \( W = mg \)

where \( g \) is the acceleration due to gravity (approximately 9.81 m/s²), we can calculate the mass of the remaining cube. Therefore:

• \( m = \frac{W}{g} = \frac{6.30}{9.81} \approx 0.642 \text{ kg} \)

With the remaining volume of the cube calculated earlier (109.29 cm³), the density \( \rho \) of the metal can be found:

• \( \rho = \frac{m}{V} = \frac{0.642}{0.10929} \approx 5.88 \text{ g/cm}^3 \)

Moreover, to know the initial weight of the cube before drilling, we consider the entire volume and multiply it by the calculated density to find the initial mass:

• \( m_{initial} = 125 \text{ cm}^3 \times 5.88 \text{ g/cm}^3 \approx 735 \text{ g} \)

Finally, converting this mass to weight involves:

• \( W_{initial} = 0.735 \text{ kg} \times 9.81 \text{ m/s}^2 \approx 7.21 \text{ N} \)

Through these calculations, we glean insights into the impact of structural modifications on weight and density.