Problem 73 - College Physics

Question

A rock is suspended from a string in air. The tension in the string is \(2.94 \mathrm{~N}\). When the rock is then dunked into a liquid and the string is allowed to go slack, it sinks and comes to rest on a spring with a spring constant of \(200 \mathrm{~N} / \mathrm{m} .\) The spring's final compression is \(1.00 \mathrm{~cm} .\) If the density of the rock is \(2500 \mathrm{~kg} / \mathrm{m}^{3},\) what is the density of the liquid?

Step-by-Step Solution

Answer

The density of the liquid is approximately 800 kg/m³.

Step 1: Understanding the physical situation

A rock is first suspended in the air, then submerged in a liquid and rests on a spring. We need to find the density of the liquid from given values: tension in the string, spring constant, and compression of the spring.

Step 2: Calculate the weight of the rock

The tension in the string when the rock is suspended is equal to the weight of the rock. Therefore, the weight of the rock, \( W \), is given by the tension in the string: \[ W = 2.94 \text{ N} \]

Step 3: Calculate the rock's mass

Use the formula for weight, \( W = mg \), to find the mass of the rock. Rearrange to find the mass: \[ m = \frac{W}{g} = \frac{2.94}{9.8} \] This gives us \( m = 0.3 \text{ kg} \).

Step 4: Find the volume of the rock

Use the density formula \( \rho = \frac{m}{V} \) to find the volume of the rock, where \( \rho = 2500 \text{ kg/m}^3 \).Rearrange to find the volume:\[ V = \frac{m}{\rho} = \frac{0.3}{2500} \]This gives \( V = 1.2 \times 10^{-4} \text{ m}^3 \).

Step 5: Calculate the force exerted by the spring

The spring is compressed by 1 cm (0.01 m) when the rock comes to rest on it. Use Hooke's Law to find the spring force, \( F_s = kx \), where \( k = 200 \text{ N/m} \) and \( x = 0.01 \text{ m} \).\[ F_s = 200 \times 0.01 = 2 \text{ N} \]

Step 6: Determine the buoyant force

When the rock is submerged, the weight of the rock is supported by both the buoyant force, \( F_b \), and the spring force, \( F_s \). Thus, \[ W = F_s + F_b \]\[ 2.94 = 2 + F_b \]Solve for \( F_b \), \[ F_b = 0.94 \text{ N} \]

Step 7: Calculate the density of the liquid

The buoyant force, \( F_b \), is equal to the weight of the liquid displaced by the rock, \( F_b = \rho_{\text{liquid}}Vg \).Rearrange to find \( \rho_{\text{liquid}}\):\[ \rho_{\text{liquid}} = \frac{F_b}{Vg} = \frac{0.94}{1.2 \times 10^{-4} \times 9.8} \]This calculation gives \( \rho_{\text{liquid}} \approx 800 \text{ kg/m}^3 \).

Key Concepts

Buoyant Force

The concept of buoyant force is central to this exercise. Whenever an object is submerged in a fluid, an upward force—known as buoyant force—is exerted by the fluid. This force is governed by Archimedes' Principle, which states that the buoyant force is equal to the weight of the fluid displaced by the object.
The magnitude of this buoyant force can be calculated using the equation:

• \( F_b = \rho_{\text{fluid}} \times V \times g \)

where \( \rho_{\text{fluid}} \) is the fluid's density, \( V \) is the volume of the submerged object, and \( g \) is the acceleration due to gravity (approximately \(9.8 \text{ m/s}^2\)).
This exercise showcases the buoyant force at work when the rock is submerged, balancing against the rock's weight and the spring force. By knowing the buoyant force, one can infer the density of the fluid if the volume of the object is known.

Density

Density is a measure of mass per unit volume and is a crucial property when analyzing buoyancy and flotation. Given as \( \rho = \frac{m}{V} \), density helps identify how much mass is contained in a given space.
In the problem, two different densities are considered:

• The density of the rock, \(2500 \ \text{kg/m}^3\), provides the rock's material compactness.
• The density of the liquid helps us understand how buoyant the liquid is when the rock is submerged.

Density plays a key role not only in defining buoyant force but also in understanding how substances interact with one another. A solid understanding of this concept allows us to solve for the fluid's density, given the buoyant force and the object's volume.

Hooke's Law

Hooke's Law describes how springs work and indicates that the force exerted by a spring is proportional to its compression or extension from its equilibrium position. The equation is:

• \( F_s = kx \)

where \( F_s \) is the spring force, \( k \) is the spring constant (in \( \text{N/m} \)), and \( x \) is the spring's displacement from its normal length.
In this problem, the rock compresses the spring by 1 cm (0.01 m) after falling into the liquid, and Hooke's Law enables us to calculate the spring force exerted—which comes out to be 2 N. By understanding the spring force, we can determine how it, alongside the buoyant force, helps support the rock’s weight when submerged.

Weight of the Rock

The weight of an object is the force with which it is pulled toward the Earth due to gravity. It can be calculated using:

• \( W = mg \)

where \( m \) is the mass of the object and \( g \) is the gravitational acceleration (approx. \(9.8 \text{ m/s}^2\)).
In this exercise, the tension in the string when suspended provides the measure of the rock's weight, which is 2.94 N. Once submerged, the weight of the rock is carried partly by the spring and partly by the buoyant force. Understanding the original weight is vital to determining how much of the weight is supported by the fluid (buoyancy) and the spring, which in turn aids in calculating the density of the liquid.