Problem 19 - University Physics with Modern Physics

Question

A spring of negligible mass has force constant \(k =\) 800 N/m. (a) How far must the spring be compressed for 1.20 J of potential energy to be stored in it? (b) You place the spring vertically with one end on the floor. You then lay a 1.60-kg book on top of the spring and release the book from rest. Find the maximum distance the spring will be compressed.

Step-by-Step Solution

Answer

Compression for 1.20 J: 0.0546 m; Max compression with book: 0.0392 m.

Step 1: Understanding the Potential Energy Formula

The potential energy stored in a compressed spring is given by the formula \( PE = \frac{1}{2} k x^2 \), where \( PE \) is the potential energy, \( k \) is the spring constant, and \( x \) is the compression distance in meters. We need to calculate \( x \) for a potential energy of 1.20 J and \( k = 800 \text{ N/m} \).

Step 2: Solving for Compression Distance (Part A)

Rearrange the formula \( PE = \frac{1}{2} k x^2 \) to solve for \( x \). This yields \( x = \sqrt{\frac{2 \times PE}{k}} \). Substituting the given values of \( PE = 1.20 \text{ J} \) and \( k = 800 \text{ N/m} \), we get \( x = \sqrt{\frac{2 \times 1.20}{800}} \). Simplify this to find \( x = \sqrt{\frac{2.4}{800}} \). Calculate to get \( x \approx 0.0546 \text{ meters} \).

Step 3: Understanding Energy Conservation for Part B

When the book is placed on the spring and released, gravitational potential energy is converted to spring potential energy at maximum compression. Set gravitational potential energy \( mgh = \frac{1}{2} k x^2 \), where \( m = 1.60 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), and \( h = x \) since the spring is compressed by height \( x \). Solve for \( x \) in \( 1.60 \times 9.8 \times x = \frac{1}{2} \times 800 \times x^2 \).

Step 4: Solving for Maximum Compression (Part B)

Simplify the energy equation: \( 15.68x = 400x^2 \). Rearrange to solve for \( x \): \( 400x^2 - 15.68x = 0 \). Factor out \( x \): \( x(400x - 15.68) = 0 \). This gives solutions \( x = 0 \) or \( 400x = 15.68 \). Solving gives \( x = \frac{15.68}{400} \). Calculate \( x \approx 0.0392 \text{ meters} \).

Step 5: Conclusion

In conclusion, the spring must be compressed approximately \( 0.0546 \text{ meters} \) for 1.20 J of potential energy, and the maximum distance it will be compressed by a 1.60 kg book is approximately \( 0.0392 \text{ meters} \).

Key Concepts

Understanding the Spring Constant

The spring constant, symbolized as \( k \), is a measure of a spring's stiffness. It's defined as the force required to stretch or compress the spring by a unit distance. In our exercise, the spring constant is given as \( k = 800 \text{ N/m} \). This means it takes 800 Newtons of force to compress or extend the spring by a single meter. The higher the spring constant, the stiffer the spring is, and the more force is required for displacement. Understanding the spring constant is crucial for solving problems related to spring potential energy as it directly influences how much energy the spring can store. In formulas, it plays a key role in determining both potential energy stored and the compression distance of the spring.

Concept of Energy Conservation

Energy conservation is a fundamental principle stating that energy cannot be created or destroyed, only transformed from one form to another. In mechanical systems, this often involves converting between kinetic, potential, and elastic (spring) energy. In the given exercise, energy conservation helps us understand how the initial gravitational potential energy (when a book is held above the spring) is transformed into spring potential energy as the spring compresses. This conservation allows us to equate the potential energy due to gravity with the elastic potential energy of the spring at maximum compression, simplifying calculations and ensuring accuracy.

Calculating Compression Distance

Compression distance, denoted by \( x \), is an essential measurement in determining how much a spring is compressed or stretched. For a given amount of potential energy, the spring's compression distance can be calculated using the formula \( PE = \frac{1}{2} k x^2 \). By rearranging this formula, we solved for \( x \) to find how far the spring must be compressed to store 1.20 Joules of energy:\[ x = \sqrt{\frac{2 \times PE}{k}} \]Using the spring constant \( k = 800 \text{ N/m} \) and \( PE = 1.20 \text{ J} \), we get \( x \approx 0.0546 \text{ meters} \). This calculation shows that a small compression distance can correspond to a significant amount of stored energy, especially in springs with high spring constants like ours.

Explaining Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field. It is given by the formula \( PE_{gravity} = mgh \), where \( m \) is the mass, \( g \) is the gravitational acceleration, and \( h \) is the height above a reference point. In the context of our exercise, when a book is placed on a vertically oriented spring, its GPE is converted to spring potential energy at maximum compression. The equation \( mgh = \frac{1}{2} k x^2 \) signifies this energy transformation. Given the book's mass of 1.60 kg, \( g = 9.8 \text{ m/s}^2 \), and equating this with the spring's energy equation, we determined the compression distance at maximum spring compression is approximately \( 0.0392 \text{ meters} \). Understanding GPE's role in this process helps clarify how potential energy is managed in mechanical systems and offers insight into designing practical applications.