College Physics
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![**Understanding the Work Done to Stretch a Spring**
The graph below illustrates the relationship between the force exerted by a spring and the distance \( x \) that the spring is stretched. This is a common physics problem that involves Hooke's Law, which states that the force \( F \) exerted by a spring is directly proportional to its extension \( x \), described by \( F = kx \), where \( k \) is the spring constant.
**Questions:**
1. How much work is needed to stretch the spring a distance of 5.0 cm, starting with it unstretched?
2. How much work is needed to stretch the spring from \( x = 2.0 \) cm to \( x = 7.0 \) cm?
**Graph Description:**
- The y-axis represents the force \( F \) in Newtons (N).
- The force ranges from 0 N to 350 N, with increments of 50 N.
- The x-axis represents the displacement \( x \) in centimeters (cm).
- The displacement ranges from 0 cm to 8 cm, with increments of 1 cm.
- The graph is a straight line passing through the origin (0,0), which indicates a linear relationship between force and displacement. This linear trend confirms Hooke's Law, \( F = kx \).
To find the work done (\( W \)) to stretch the spring, we use the formula for the work done in stretching a linear spring:
\[ W = \frac{1}{2} k x^2 \]
**Solving the Questions:**
1. **For stretching the spring from 0 cm to 5.0 cm:**
- Determine the spring constant \( k \):
By reading from the graph, at \( x = 5.0 \) cm, \( F = 200 \) N.
Using \( F = kx \):
\[
200 \, \text{N} = k \cdot 5.0 \, \text{cm}
\]
Convert 5.0 cm to meters:
\[
5.0 \, \text{cm} = 0.05 \, \text{m}
\]
Then:
\[
k = \frac{200 \, \text{N}}{0.05 \, \text{m](https://content.bartleby.com/qna-images/question/853bd72d-1b9a-40ff-a061-0c00f168c5bc/eb3f0324-a5fb-42db-ad5a-a759999cfc34/46b5q5an_processed.png)
Transcribed Image Text:**Understanding the Work Done to Stretch a Spring**
The graph below illustrates the relationship between the force exerted by a spring and the distance \( x \) that the spring is stretched. This is a common physics problem that involves Hooke's Law, which states that the force \( F \) exerted by a spring is directly proportional to its extension \( x \), described by \( F = kx \), where \( k \) is the spring constant.
**Questions:**
1. How much work is needed to stretch the spring a distance of 5.0 cm, starting with it unstretched?
2. How much work is needed to stretch the spring from \( x = 2.0 \) cm to \( x = 7.0 \) cm?
**Graph Description:**
- The y-axis represents the force \( F \) in Newtons (N).
- The force ranges from 0 N to 350 N, with increments of 50 N.
- The x-axis represents the displacement \( x \) in centimeters (cm).
- The displacement ranges from 0 cm to 8 cm, with increments of 1 cm.
- The graph is a straight line passing through the origin (0,0), which indicates a linear relationship between force and displacement. This linear trend confirms Hooke's Law, \( F = kx \).
To find the work done (\( W \)) to stretch the spring, we use the formula for the work done in stretching a linear spring:
\[ W = \frac{1}{2} k x^2 \]
**Solving the Questions:**
1. **For stretching the spring from 0 cm to 5.0 cm:**
- Determine the spring constant \( k \):
By reading from the graph, at \( x = 5.0 \) cm, \( F = 200 \) N.
Using \( F = kx \):
\[
200 \, \text{N} = k \cdot 5.0 \, \text{cm}
\]
Convert 5.0 cm to meters:
\[
5.0 \, \text{cm} = 0.05 \, \text{m}
\]
Then:
\[
k = \frac{200 \, \text{N}}{0.05 \, \text{m
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