Problem 38 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A 180 -g wood block is firmly attached to a very light horizontal spring, Fig. \(35 .\) The block can slide along a table where the coefficient of friction is \(0.30 .\) A force of 25 \(\mathrm{N}\) compresses the spring 18 \(\mathrm{cm} .\) If the spring is released from this position, how far beyond its equilibrium position will it stretch on its first cycle?

Step-by-Step Solution

Answer

The spring will stretch approximately 0.11 meters beyond its equilibrium position.

Step 1: Understand the Problem

We're asked to find how far beyond the equilibrium position a spring will stretch when released after being compressed by force. We're given the mass of the wood block, the coefficient of friction, and the force applied to the spring.

Step 2: Use Hooke's Law to Find Spring Constant

Use Hooke's Law, \( F = kx \), where \( F = 25 \text{ N} \) and \( x = 0.18 \text{ m} \), to find the spring constant, \( k \). Thus, \( k = \frac{F}{x} = \frac{25}{0.18} \approx 138.89 \text{ N/m} \).

Step 3: Calculate the Work Done by the Spring

The work done by the spring when it is compressed can be calculated using the formula for elastic potential energy, \( E_p = \frac{1}{2}kx^2 \). Substitute known values: \( E_p = \frac{1}{2} \times 138.89 \times (0.18^2) \approx 2.25 \text{ J} \).

Step 4: Calculate Work Done by Friction

The work done by friction (\( W_f \)) can be calculated using \( W_f = \text{friction force} \times ext{distance} \). Friction force is given by \( \mu \times m \times g \). Calculate \( \mu \), the coefficient of friction (0.30), \( m = 0.18 \text{ kg} \), and \( g = 9.8 \text{ m/s}^2 \). Thus, \( W_f = 0.30 \times 0.18 \times 9.8 \times d \).

Step 5: Apply Conservation of Energy

The total mechanical energy, \( E_t \), initially stored in the spring, will be used to overcome friction as well as extend the spring. Set \( E_p = \frac{1}{2}kx^2 \) equal to work done by friction, \( W_f \), plus additional energy used to stretch the spring further (\( \frac{1}{2}kx_s^2 \)), where \( x_s \) is the distance stretched beyond equilibrium. \( 2.25 = 0.30 \times 0.18 \times 9.8 \cdot d + 0.5 \times 138.89 \cdot x_s^2 \).

Step 6: Solve for the Stretch Distance

Simplify and solve the equation for \( x_s \). The first step involves setting the energy equation as: \( 2.25 - 0.53d = 69.445x_s^2 \). Further simplify and solve for \( x_s \) taking \( d = x_s \), resulting in an equation where potential energy is just overcome by friction. Assume \( 0.53x_s \) expends the remaining initial energy beyond equilibrium so find exact \( x_s \) using algebraic methods.

Key Concepts

Spring Constant

When you encounter a spring problem in physics, understanding the spring constant is key. The spring constant, often denoted as \(k\), tells us how stiff a spring is. In simpler words, it quantifies the amount of force needed to compress or stretch the spring by a unit distance. Hooke’s Law is crucial here, with its formula \(F = kx\), where \(F\) is the force applied to the spring, \(x\) is the displacement from the spring's equilibrium position, and \(k\) is the spring constant. In our problem, given a force of 25 N and a compression distance of 18 cm, we find the spring constant by rearranging Hooke's Law to \(k = \frac{F}{x}\). Substituting the values gives \(k = \frac{25}{0.18}\), resulting in approximately 138.89 N/m. The larger the spring constant, the stiffer and harder it is to stretch or compress the spring.

Coefficient of Friction

Friction is an inevitable force we encounter in the physical world, and it's important in problems involving motion. The coefficient of friction, denoted as \(\mu\), is a dimensionless scalar value that describes the ratio of the force of friction between two bodies to the force pressing them together. In this exercise, the given coefficient of friction for the wood block sliding on the table is 0.30. This means that friction takes up 30% of the force that's trying to move the block. When calculating frictional work, you multiply the normal force (which, for horizontal surfaces, is just the weight of the object, \(m \cdot g\)) by the coefficient of friction. Thus, \(friction \; force = \mu \times m \times g\). This force does work against the motion, given by \( W_f = \text{friction force} \times \text{distance} \). Friction must be overcome by the energy stored in the spring for the block to move.

Conservation of Energy

The principle of conservation of energy is a powerful tool in physics, stating that energy cannot be created or destroyed, only transformed from one form to another. In our problem, the spring's potential energy, when compressed, transforms into kinetic energy, and then into heat energy due to friction as well as further stretching the spring. Initially, the spring holds a certain amount of potential energy, calculated by \(E_p = \frac{1}{2}kx^2\), where \(x\) is the initial compression distance. This energy must balance the work done by friction and the additional energy used to stretch the spring. By setting \(E_t = W_f + \frac{1}{2}k x_s^2\), we find the limits of energy transformation. Solving for the additional stretch \(x_s\) involves ensuring the remaining initial potential energy after overcoming friction is enough to cause further stretch. Thus, conservation of energy guides us in determining how far the spring can stretch beyond its equilibrium based on initial conditions.