Chapter 11: Problem 39
A 500 -g object is attached to the end of an initially unstretched vertical spring for which \(k=30 \mathrm{~N} / \mathrm{m}\). The object is then released, so that it falls and stretches the spring. How far will it fall before stopping? [Hint: The \(\mathrm{PE}_{G}\) lost by the falling object must appear as \(\left.\mathrm{PE}_{e} .\right]\)
Short Answer
Step by step solution
Understand the Problem
Calculate Gravitational Potential Energy Lost
Calculate Elastic Potential Energy Gained
Set Up the Energy Balance Equation
Solve for x
Compute Final Value of x
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Potential Energy
- \( \mathrm{PE_G} = m \cdot g \cdot h \)
- \( m \) is the mass of the object (in kilograms)
- \( g \) is the acceleration due to gravity (approximately 9.8 m/s² on Earth)
- \( h \) is the height (in meters) above the reference point
Elastic Potential Energy
- The formula for elastic potential energy (\( \mathrm{PE_e} \)) is given by:
- \( \mathrm{PE_e} = \frac{1}{2} k x^2 \)
- \( k \) represents the spring constant (in N/m), measuring the stiffness of the spring.
- \( x \) is the displacement from the original unstressed position (in meters).
Spring Constant
- The unit of the spring constant is Newtons per meter (N/m).
- A higher spring constant means a stiffer spring.
- Equation context: \( F = k x \), where \( F \) is the force applied, and \( x \) is the displacement.