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The graph shows the x-directed force F, acting on an object as a function of the position x of the object. For each numbered interval given, find the work W, done on the object. 1) from x = 0 m to x = 2.50 m W₁ = 2) from x 3.00 m to x = 5.90 m W₂ = 9 W3 = 0 Incorrect 3) from x = 7.00 m to x = 9.10 m -1 Incorrect J J F (N) 0 -4- 6 7 8 9 10 x(m) < Feedback Macmillan Learning Consider the interval from x to x = 5.90 m. This interval can be broken in parts. The first part is the trian area above the axis, the secon triangular area below the axis third is a rectangular area bel the axis. The total work for this interva sum of all of the contribution- represented by these areas. Remember that areas below th are negative.

The graph shows the x-directed force F, acting on an object as a function of the position x of the object. For each numbered interval given, find the work W, done on the object. 1) from x = 0 m to x = 2.50 m W₁ = 2) from x 3.00 m to x = 5.90 m W₂ = 9 W3 = 0 Incorrect 3) from x = 7.00 m to x = 9.10 m -1 Incorrect J J F (N) 0 -4- 6 7 8 9 10 x(m) < Feedback Macmillan Learning Consider the interval from x to x = 5.90 m. This interval can be broken in parts. The first part is the trian area above the axis, the secon triangular area below the axis third is a rectangular area bel the axis. The total work for this interva sum of all of the contribution- represented by these areas. Remember that areas below th are negative.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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The graph shows the x-directed force F, acting on an object as a function of the position x of the object. For each numbered interval given, find the work W; done on the object. 1) from x = 0 m to x = 2.50 m W₁ = 9 2) from x = 3.00 m to x = 5.90 m W₂ = 0 W₁ = Incorrect 3) from x = 7.00 m to x = 9.10 m -1 Incorrect J J J 6 4 2 0 -2 -4 -6 2 3 4 5 6 7 8 9 10 x(m) < Feedback Ⓒ Macmillan Learning Consider the interval from x = 3.00 m to x = 5.90 m. This interval can be broken into three parts. The first part is the triangular area above the axis, the second is a triangular area below the axis, and the third is a rectangular area below the axis. The total work for this interval is the sum of all of the contributions represented by these areas. Remember that areas below the x-axis are negative.
Transcribed Image Text:The graph shows the x-directed force F, acting on an object as a function of the position x of the object. For each numbered interval given, find the work W; done on the object. 1) from x = 0 m to x = 2.50 m W₁ = 9 2) from x = 3.00 m to x = 5.90 m W₂ = 0 W₁ = Incorrect 3) from x = 7.00 m to x = 9.10 m -1 Incorrect J J J 6 4 2 0 -2 -4 -6 2 3 4 5 6 7 8 9 10 x(m) < Feedback Ⓒ Macmillan Learning Consider the interval from x = 3.00 m to x = 5.90 m. This interval can be broken into three parts. The first part is the triangular area above the axis, the second is a triangular area below the axis, and the third is a rectangular area below the axis. The total work for this interval is the sum of all of the contributions represented by these areas. Remember that areas below the x-axis are negative.
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The force in units of newtons is plotted along y axis and the position in units of meter is plotted along x axis 

 

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