Problem 3 - Physics

Question

A \(0.70\) -kg block is hung from and stretches a spring that is attached to the ceiling. A second block is attached to the first one, and the amount that the spring stretches from its unstrained length triples. What is the mass of the second block?

Step-by-Step Solution

Answer

The mass of the second block is 1.40 kg.

Step 1: Understanding Hooke's Law

Hooke's Law states that the force exerted by a spring is proportional to its extension (or compression): \( F = kx \), where \( k \) is the spring constant and \( x \) is the extension. For the first block, the force is \( F_1 = m_1g \), where \( m_1 = 0.70 \) kg is the mass and \( g \) is the gravitational acceleration.

Step 2: Initial Extension Calculation

The initial extension \( x_1 \) caused by the first block can be described by \( F_1 = kx_1 = m_1g \). Thus, \( kx_1 = 0.70 \cdot g \).

Step 3: Triple Extension with Two Blocks

When the second block is added, the extension triples: \( 3x_1 \). The new force \( F_2 \) is the total weight of both blocks, \( F_2 = (m_1 + m_2)g \). So, \( k(3x_1) = (0.70 + m_2)g \).

Step 4: Equating Forces and Solving for \( m_2 \)

Since \( kx_1 = 0.70 \cdot g \), substitute \( kx_1 \) into the equation: \( 3(0.70 \cdot g) = (0.70 + m_2)g \). Simplifying gives \( 2.10g = 0.70g + m_2g \). Dividing each term by \( g \), we find \( 2.10 = 0.70 + m_2 \).

Step 5: Calculate Mass of Second Block

Solving for \( m_2 \), we have \( m_2 = 2.10 - 0.70 = 1.40 \) kg.

Key Concepts

Understanding the Spring Constant

The spring constant, denoted as \( k \), is a pivotal element in understanding how springs behave under different forces. It measures the stiffness or rigidity of a spring. Specifically, it's the amount of force required to stretch or compress a spring by a unit length. A higher spring constant means a stiffer spring, requiring more force to achieve the same extension.

In our exercise, the spring constant allows us to connect the force exerted by the spring back to Hooke's Law: \( F = kx \). Here, \( F \) represents the force exerted, \( k \) is the spring constant, and \( x \) is the extension of the spring. Understanding this relationship helps determine how much force is at play when a mass is attached to a spring.

Calculating the Extension of a Spring

The extension of a spring refers to how much the spring stretches or compresses under force. In the context of Hooke’s Law, extension \( x \) is a crucial variable that comes into play when calculating the force exerted by the spring.

For the initial block of mass \( 0.70 \, \text{kg}\), the extension is given as \( x_1 \). Using Hooke's Law, this initial extension can be computed using the formula \( kx_1 = m_1g \), where \( m_1 \) is the mass, and \( g \) is the gravitational force. As more weight is added, the extension becomes \( 3x_1 \), which indicates the spring's stretch triples under the added load.

The Role of Gravitational Force

Gravitational force acts on any mass near the surface of the Earth, pulling it downward. This force is calculated by \( F_g = mg \), where \( m \) is the mass of the object, and \( g \) is the gravitational acceleration, approximately \( 9.8 \, \text{m/s}^2 \).

In the problem, each block experiences a gravitational force that contributes to the total force exerted on the spring. Initially, the gravitational force on the first block is \( 0.70g \). When the second block is attached, the combined gravitational force that stretches the spring is \((m_1 + m_2)g\). Understanding gravitational force is key in computing the total force acting on the spring and subsequently finding how much it stretches.

Mass Calculation of the Second Block

The core of this problem lies in calculating the mass of the second block, \( m_2 \). Once the spring's total stretch triples with the addition of the second block, and using the relationship from Hooke's Law, we express the new force as \( k(3x_1) = (0.70 + m_2)g \).

By simplifying and substituting values, we find \( 3(0.70g) = (0.70 + m_2)g \). This equation allows us to solve for \( m_2 \) directly, yielding \( m_2 = 1.40 \, \text{kg} \). Thus, determining the mass of the second block becomes an exercise in understanding how the extension of the spring relates to the force and the masses involved.