Problem 37 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A 0.620 -kg wood block is firmly attached to a very light horizontal spring \((k=180 \mathrm{~N} / \mathrm{m})\) as shown in Fig. \(8-35 .\) This block-spring system, when compressed \(5.0 \mathrm{~cm}\) and released, stretches out \(2.3 \mathrm{~cm}\) beyond the equilibrium position before stopping and turning back. What is the coefficient of kinetic friction between the block and the table?

Step-by-Step Solution

Answer

The coefficient of kinetic friction is approximately 0.139.

Step 1: Understand the Problem

We have a block-spring system where a block is attached to a spring with a spring constant, \(k = 180 \, \text{N/m}\). The block is compressed by \(5.0 \, \text{cm}\) and then released, stretching past the equilibrium position by \(2.3 \, \text{cm}\). We need to find the coefficient of kinetic friction between the block and the surface.

Step 2: Convert Distance Measurements

Convert the distances from centimeters to meters. The initial compression of the spring is \(5.0 \, \text{cm} = 0.050 \, \text{m}\). The distance the spring stretches beyond the equilibrium position is \(2.3 \, \text{cm} = 0.023 \, \text{m}\).

Step 3: Calculate the Work Done by Friction

The initial potential energy stored in the spring is given by \(\frac{1}{2} k x_i^2\), with \(x_i = 0.050 \, \text{m}\). The work done by friction, \(W_f\), equals the change in mechanical energy of the system. The final energy is stored in the spring when the stretch is \(0.023 \, \text{m}\), given by \(\frac{1}{2} k x_f^2\).

Step 4: Write the Energy Balance Equation

The energy lost to friction is the difference between initial and final potential energy: \[\frac{1}{2} k x_i^2 - \frac{1}{2} k x_f^2 = f_k \cdot d\], where \(f_k\) is the kinetic friction force, and \(d = x_i + x_f\) is the total distance traveled.

Step 5: Calculate the Kinetic Friction Force

The kinetic friction force is expressed as \(f_k = \mu_k mg\), where \(\mu_k\) is the coefficient of kinetic friction, \(m = 0.620 \, \text{kg}\) is the mass, and \(g = 9.8 \, \text{m/s}^2\) is the acceleration due to gravity. Substitute \(f_k = \mu_k mg\) into the energy balance equation.

Step 6: Solve for the Coefficient of Kinetic Friction

Plug in the values: initial compression \(x_i = 0.050 \, \text{m}\), final stretch \(x_f = 0.023 \, \text{m}\), \(k = 180 \, \text{N/m}\), \(m = 0.620 \, \text{kg}\), and \(g = 9.8 \, \text{m/s}^2\), and solve for \(\mu_k\) using the equation: \[\mu_k = \frac{\frac{1}{2} k (x_i^2 - x_f^2)}{mg \cdot (x_i + x_f)}\].

Key Concepts

Hooke's Law

Hooke's Law is a fundamental principle in physics that describes the behavior of springs. It states that the force required to compress or extend a spring by a certain distance is directly proportional to that distance. The mathematical expression of Hooke's Law is:\[ F = -kx \]where:

• \( F \) is the force applied to the spring (in Newtons, N)
• \( k \) is the spring constant (in N/m), a measure of the spring's stiffness
• \( x \) is the displacement from the spring's equilibrium position (in meters, m)

This principle is crucial for solving problems involving springs, such as the movement of the block attached to a spring in our exercise. Understanding Hooke's Law helps us predict how much energy is stored in a spring and how much force is exerted by it during compression and extension.

Spring Constant

The spring constant, denoted by \( k \), is a key parameter in Hooke's Law, representing the stiffness of a spring. A larger spring constant indicates a stiffer spring, which requires more force to compress or extend.In our problem, the spring constant is given as \( 180 \, \text{N/m} \). This means that for every meter the spring is stretched or compressed from its equilibrium position, it exerts a force of 180 Newtons.A notable feature of the spring constant is its role in the potential energy stored within the spring. A stiffer spring (higher \( k \)) will store more potential energy for a given displacement. This concept is essential when calculating how much energy is lost or transformed in mechanical systems involving springs. In the case of our exercise, understanding the spring constant helps us determine the initial and final potential energy stored by the spring.

Potential Energy

Potential energy in the context of a spring is the energy stored due to its position or condition. When dealing with springs, this energy originates from the displacement of the spring from its equilibrium position. The equation for calculating spring potential energy is:\[ PE_{spring} = \frac{1}{2} k x^2 \]where:

• \( PE_{spring} \) is potential energy of the spring (in Joules, J)
• \( k \) is the spring constant (in N/m)
• \( x \) is the displacement from equilibrium (in meters, m)

In the exercise problem, the block compresses the spring initially and eventually stretches it, leading to changes in potential energy. Calculating these energy changes is crucial to understanding the energy transfer in the system and the work done against friction.

Friction in Mechanics

Friction is a force opposing the motion of objects sliding against each other. In the context of mechanics and our exercise, we are interested in kinetic friction, which acts between the block and the surface as it moves.The kinetic friction force, \( f_k \), is calculated as:\[ f_k = \mu_k mg \]where:

• \( \mu_k \) is the coefficient of kinetic friction (dimensionless)
• \( m \) is the mass of the object (in kilograms, kg)
• \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \))

The coefficient of kinetic friction, \( \mu_k \), is a measure of how much force is needed to move one object over another. In our exercise, calculating \( \mu_k \) involves analyzing the energy changes within the spring system and equating them to the work done by friction. This helps us understand the interplay between energy stored in the spring and energy dissipated through friction.