How much work is done. on jhe precticle when it reaches x = fro when it reeches x
How much work is done. on jhe precticle when it reaches x = fro when it reeches x
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![### Educational Content on Work Done by a Force on a Particle
**Problem Statement:**
A particle moving along the x-axis experiences the force shown in the graph. How much work is done on the particle when it reaches \( x = 4.0 \, \text{m} \)?
**Graph Description:**
- The graph plots force \( F_x \) (in Newtons) on the y-axis against position \( x \) (in meters) on the x-axis.
- The data points on the graph are as follows:
- \( x = 0 \, \text{m} \), \( F_x = 0 \, \text{N} \)
- \( x = 2 \, \text{m} \), \( F_x = 2 \, \text{N} \)
- \( x = 4 \, \text{m} \), \( F_x = 4 \, \text{N} \)
- The graph illustrates a linear increase in force with respect to position.
**Concept Explanation:**
The work done by a force on a particle along the x-axis can be calculated using the integral of the force with respect to displacement. For this linear force graph, the area under the line between the points indicates the work done:
- From \( x = 0 \) to \( x = 2 \), the area is a right triangle with a base of 2 m and height of 2 N.
- From \( x = 2 \) to \( x = 4 \), the area is another right triangle with a base of 2 m and height of 2 N.
By calculating these triangles' areas, you can find the total work done on the particle from \( x = 0 \, \text{m} \) to \( x = 4 \, \text{m} \).
**Calculation:**
The work \( W \) done is equal to the area under the force vs. position graph:
- **Area of first triangle**:
\[
\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \, \text{m} \times 2 \, \text{N} = 2 \, \text{J}
\]
- **Area of second triangle**:
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Transcribed Image Text:### Educational Content on Work Done by a Force on a Particle
**Problem Statement:**
A particle moving along the x-axis experiences the force shown in the graph. How much work is done on the particle when it reaches \( x = 4.0 \, \text{m} \)?
**Graph Description:**
- The graph plots force \( F_x \) (in Newtons) on the y-axis against position \( x \) (in meters) on the x-axis.
- The data points on the graph are as follows:
- \( x = 0 \, \text{m} \), \( F_x = 0 \, \text{N} \)
- \( x = 2 \, \text{m} \), \( F_x = 2 \, \text{N} \)
- \( x = 4 \, \text{m} \), \( F_x = 4 \, \text{N} \)
- The graph illustrates a linear increase in force with respect to position.
**Concept Explanation:**
The work done by a force on a particle along the x-axis can be calculated using the integral of the force with respect to displacement. For this linear force graph, the area under the line between the points indicates the work done:
- From \( x = 0 \) to \( x = 2 \), the area is a right triangle with a base of 2 m and height of 2 N.
- From \( x = 2 \) to \( x = 4 \), the area is another right triangle with a base of 2 m and height of 2 N.
By calculating these triangles' areas, you can find the total work done on the particle from \( x = 0 \, \text{m} \) to \( x = 4 \, \text{m} \).
**Calculation:**
The work \( W \) done is equal to the area under the force vs. position graph:
- **Area of first triangle**:
\[
\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \, \text{m} \times 2 \, \text{N} = 2 \, \text{J}
\]
- **Area of second triangle**:
\
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