Chapter 5: Problem 47
Find the work done in extending a spring with \(k=150 \mathrm{~N} / \mathrm{m}\) from \(x=0.10 \mathrm{~m}\) to \(x=0.30 \mathrm{~m}\).
Short Answer
Expert verified
The work done is 6 joules.
Step by step solution
01
Understand Hooke's Law
Work done on a spring can be calculated using Hooke's Law, which states that the force required to extend or compress a spring by some distance \( x \) from its natural length is proportional to \( x \). The formula is: \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement.
02
Recall the Work Done Formula
The work done on a spring is calculated using the formula: \( W = \frac{1}{2}k(x_f^2 - x_i^2) \). Here, \( W \) is the work done, \( k \) is the spring constant, \( x_i \) is the initial displacement, and \( x_f \) is the final displacement.
03
Identify Given Values
From the problem, we know that \( k = 150 \, \text{N/m} \), \( x_i = 0.10 \, \text{m} \), and \( x_f = 0.30 \, \text{m} \). These are the values that will be substituted into the work done formula.
04
Substitute Values into Work Formula
Substitute the given values into the work formula: \( W = \frac{1}{2} \times 150 \times (0.30^2 - 0.10^2) \).
05
Calculate Inside the Parentheses
First, calculate \( 0.30^2 - 0.10^2 \): \[ 0.30^2 = 0.09 \, \text{m}^2 \]\[ 0.10^2 = 0.01 \, \text{m}^2 \]\[ 0.30^2 - 0.10^2 = 0.09 - 0.01 = 0.08 \, \text{m}^2 \]
06
Calculate the Work Done
Now substitute back into the formula to find the work done: \[ W = \frac{1}{2} \times 150 \times 0.08 \]\[ W = 75 \times 0.08 = 6 \, \text{J} \].
07
Finalize the Answer
The work done in extending the spring from 0.10 m to 0.30 m is 6 joules.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Hooke's Law
Hooke's Law is a fundamental concept in physics that provides insight into how springs behave. It describes the relationship between the force applied to a spring and the resulting displacement or stretch from its natural length. Simply put, when you pull or push a spring, it resists with a force that is directly proportional to the amount you stretch or compress it. This relationship is described by the formula:
\[ F = kx \]
\[ F = kx \]
- \( F \) is the force applied on the spring (in newtons, N).
- \( k \) is the spring constant, a measure of the spring's stiffness (in newtons per meter, N/m).
- \( x \) is the displacement of the spring from its equilibrium position (in meters, m).
Spring Constant
The spring constant, denoted by \( k \), is a crucial parameter in understanding a spring's characteristics. It tells us the amount of force needed to stretch or compress a spring by one unit of length. It also indicates how stiff or flexible a spring is.
For example, in the exercise provided, the spring has a spring constant of 150 N/m. This means if you were to stretch or compress the spring by 1 meter from its natural length, it would require a force of 150 newtons.
Here are some essential points about the spring constant:
For example, in the exercise provided, the spring has a spring constant of 150 N/m. This means if you were to stretch or compress the spring by 1 meter from its natural length, it would require a force of 150 newtons.
Here are some essential points about the spring constant:
- A larger spring constant signifies a stiffer spring that requires more force to stretch.
- A smaller spring constant means a more flexible spring that requires less force for the same amount of stretch.
- The spring constant is unique to each spring and depends on its material and structure.
Work Done Formula
The work done in stretching or compressing a spring is a critical measure of energy transfer. This is quantified by the work done formula in the context of springs which is derived from integrating Hooke's Law. The formula is:
\[ W = \frac{1}{2}k(x_f^2 - x_i^2) \]
In this formula:
\[ W = \frac{1}{2}k(x_f^2 - x_i^2) \]
In this formula:
- \( W \) is the work done on the spring, calculated in joules (J).
- \( x_i \) and \( x_f \) are the initial and final displacements of the spring from its rest position, respectively.
- \( k \) remains the spring constant, which indicates the inherent stiffness.
- The initial displacement \( x_i = 0.10 \, \text{m} \) and final displacement \( x_f = 0.30 \, \text{m} \)
- Substituting these into the formula: \[ W = \frac{1}{2} \times 150 \times (0.30^2 - 0.10^2) \] results in 6 joules.