Why is the coefficient two in the work done by friction equation?

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The discussion centers on the confusion regarding the coefficient of two in the work done by friction equation. Participants clarify that the total distance traveled by the block in one cycle exceeds the simple sum of initial and final positions, leading to the multiplication by two. The distinction between the first and second edition manuals is highlighted, particularly in how they define the positions xi and xf. In the second edition, these terms represent different offsets, contributing to the total distance calculation. Understanding this context resolves the confusion about why the work done by friction includes the coefficient two.
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Homework Statement
below
Relevant Equations
na
1611450520083.png

Solution in 2ed manual:
1611450607765.png

Solution in 1ed manual:
1611450680206.png

Could someone explain why the work done by friction is multiplied by 2? I get that the distance traveled by the block in one cycle is greater than ##x_f+x_i##, but why is the coefficient two?
 
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Leo Liu said:
Homework Statement:: below
Relevant Equations:: na

View attachment 276734
Solution in 2ed manual:
View attachment 276735
Solution in 1ed manual:
View attachment 276736
Could someone explain why the work done by friction is multiplied by 2? I get that the distance traveled by the block in one cycle is greater than ##x_f+x_i##, but why is the coefficient two?
It is confusing that the two solutions use xi and xf to mean different things.
In the second one (1ed), they are both offsets from the relaxed position instead of from the wall, and xf means the new rightmost extension after one cycle.
That cycle involves traveling xi to the left to reach xo, then another (xi+xf)/2 to the left for its first leftmost position, then back through xo to its new rightmost position.
Total distance, 2(xi+xf).
 
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