3. A graph of the force on an object as a function of its position is shown in the figure. Determine the amount of work done by this force on the object during a displacement from x =-2.00 m to x 2.00 m. (Assume an accuracy of 3 significant figures for the numbers on the graph.) F(N) 4+ -54 x(m)

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Chapter1: Units, Trigonometry. And Vectors
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### Problem Statement

3. A graph of the force on an object as a function of its position is shown in the figure. Determine the amount of work done by this force on the object during a displacement from \( x = -2.00 \, \text{m} \) to \( x = 2.00 \, \text{m} \). (Assume an accuracy of 3 significant figures for the numbers on the graph.)

### Graph Description

The graph displays the force \( F \) in newtons (N) on the vertical axis versus the position \( x \) in meters (m) on the horizontal axis. The graph is a straight line with a negative slope, beginning at \( x = -5 \, \text{m} \) with \( F = 5 \, \text{N} \) and decreasing to \( x = 5 \, \text{m} \) where \( F = -5 \, \text{N} \). The line crosses the horizontal axis at \( F = 0 \) when \( x = 0 \, \text{m} \).

To determine the work done, calculate the area under the force-position graph from \( x = -2.00 \, \text{m} \) to \( x = 2.00 \, \text{m} \). Since the graph is a straight line, this area can be split into geometric shapes (a triangle and a rectangle) to simplify the calculation.
Transcribed Image Text:### Problem Statement 3. A graph of the force on an object as a function of its position is shown in the figure. Determine the amount of work done by this force on the object during a displacement from \( x = -2.00 \, \text{m} \) to \( x = 2.00 \, \text{m} \). (Assume an accuracy of 3 significant figures for the numbers on the graph.) ### Graph Description The graph displays the force \( F \) in newtons (N) on the vertical axis versus the position \( x \) in meters (m) on the horizontal axis. The graph is a straight line with a negative slope, beginning at \( x = -5 \, \text{m} \) with \( F = 5 \, \text{N} \) and decreasing to \( x = 5 \, \text{m} \) where \( F = -5 \, \text{N} \). The line crosses the horizontal axis at \( F = 0 \) when \( x = 0 \, \text{m} \). To determine the work done, calculate the area under the force-position graph from \( x = -2.00 \, \text{m} \) to \( x = 2.00 \, \text{m} \). Since the graph is a straight line, this area can be split into geometric shapes (a triangle and a rectangle) to simplify the calculation.
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