The graph in the figure below shows the magnitude of the force ex- erted by a given spring as a function of the distance x the spring is stretched. How much work is needed to stretch this spring: (a) a distance of 5.0 cm, starting with it unstretched, and (b) from x = = 2.0 cm to x = 7.0 cm? F (N) 350 300 250 200 150

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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
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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