Problem 66 - University Physics with Modern Physics

Question

A basket of negligible weight hangs from a vertical spring scale of force constant 1500 N/m. (a) If you suddenly put a 3.0-kg adobe brick in the basket, find the maximum distance that the spring will stretch. (b) If, instead, you release the brick from 1.0 m above the basket, by how much will the spring stretch at its maximum elongation?

Step-by-Step Solution

Answer

(a) 0.0196 m; (b) About 0.198 m

Step 1: Identify the System Parameters

We need to focus on the spring constant (k) and the mass of the brick (m). The spring constant is given as \( k = 1500 \, \text{N/m} \). The mass of the adobe brick is \( m = 3.0 \, \text{kg} \).

Step 2: Determine the Max Spring Stretch (Static Situation)

We apply Hooke's Law, \( F = kx \), where \( F \) is the force by the weight of the brick and \( x \) is the displacement. The force here is simply \( F = mg \), where \( g = 9.8 \, \text{m/s}^2 \). Hence, \( x = \frac{mg}{k} = \frac{3.0 \times 9.8}{1500} \approx 0.0196 \, \text{m} \). This is the distance the spring stretches when the brick is gently placed in the basket.

Step 3: Determine the Max Spring Stretch (Dynamic Situation)

We use energy considerations to solve this. When the brick is released from 1.0 m above, the gravitational potential energy becomes elastic potential energy: \( mgh = \frac{1}{2}kx^2 \). Substituting values, \( 3.0 \times 9.8 \times 1.0 = \frac{1}{2} \times 1500 \times x^2 \). Solving, \( x^2 = \frac{3.0 \times 9.8 \times 1.0 \times 2}{1500} \approx 0.0392 \). Thus, \( x \approx 0.198 \, \text{m} \).

Step 4: Final Calculations and Checks

Recalculate to ensure values are logical and match physics principles. When the brick is released from 1.0 m, ensure the energy conversion formula balances. Both scenarios result in valid spring stretches calculated from theoretical principles.

Key Concepts

Spring Constant

The spring constant, often symbolized as \( k \), is a fundamental concept in understanding how springs behave under force. It measures the stiffness of a spring, determining how resistant it is to deformation. When you apply a force to a spring, the spring constant tells you how much it will stretch or compress. The larger the spring constant, the stiffer the spring and the more force needed to achieve a given change in length.
For example, in our exercise, the spring constant is given as 1500 N/m. This means it requires 1500 Newtons of force to stretch or compress the spring by 1 meter. We use Hooke's Law, \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement, to find how much a spring will stretch under a given weight.

• A larger spring constant implies a stiffer spring.
• It directly relates force and displacement through Hooke's Law.

Gravitational Potential Energy

Gravitational potential energy is the energy stored due to an object's position above the ground. This energy depends on the object's mass, height, and gravity. The formula for gravitational potential energy is \( U = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity (9.8 \text{m/s}^2), and \( h \) is the height above the reference point.
In the exercise, when the adobe brick is held 1 meter above the basket, it holds gravitational potential energy. As it falls, this energy transforms, influencing how far the spring stretches. Understanding gravitational potential energy helps to explain how height and weight impact the energy involved when an object moves under the influence of gravity.

• Energy depends on the position relative to the ground.
• Key in calculating energy conversion in falling objects.

Elastic Potential Energy

Elastic potential energy is energy stored within a system when it is deformed. For a spring, this refers to the energy stored when the spring is compressed or stretched. The formula to calculate elastic potential energy is \( U = \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the displacement from the spring's equilibrium position.
In our exercise, when the brick is released and stretches the spring, this potential energy comes from the spring as it elongates under the load. Understanding this concept is crucial to solving problems where deformation occurs, and it explains how the spring's potential energy grows with increased stretching.

• Directly related to the work done to deform the spring.
• Increases with greater displacement and spring stiffness.

Energy Conversion

Energy conversion refers to the process of transforming energy from one form to another. In the context of the exercise, this relates to converting gravitational potential energy into elastic potential energy. Initially, the brick at a height possesses gravitational potential energy. Upon release, this energy converts, stretching the spring and turning into elastic potential energy.
In solving such problems, we equate the initial potential energy to the spring's potential energy at maximum stretch, using the principles of energy conservation. Recognizing energy conversion's role demonstrates how different energy types maintain equilibrium, aiding in prediction and analysis.

• Ensures compliance with energy conservation laws.
• Allows predictive calculations for system behavior changes.

Physics Problem Solving

Physics problem solving involves a systematic approach to exploring physical phenomena and finding solutions to real-life situations. This exercise demonstrates key problem-solving steps: identifying parameters, applying appropriate physical laws, and evaluating results.
Steps include understanding given data (spring constant, mass), applying Hooke's Law or energy equations, and checking results to ensure they adhere to physical laws. Effective problem-solving combines theory with practical calculation checks, ensuring logical, realistic outcomes.

• Begins with identifying known and unknown parameters.
• Utilizes fundamental physics laws for accurate analysis.
• Requires validation of results to ensure realistic solutions.