How Does Friction Impact Work Calculation in Physics?

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Friction impacts work calculation in physics by always acting in the opposite direction of displacement, resulting in a negative work value. The calculations for work done by friction can be derived using integral methods or kinetic energy changes, leading to a consistent result of approximately -661J. The importance of significant figures is highlighted, particularly with the example of a 68kg mass, which can affect the final answer. It is emphasized that sharing solutions and methods is beneficial for others learning physics, especially regarding the negative nature of work done by friction. Posting solutions can aid in understanding and provide instructional value for future learners.
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This guy is stealing my bases!

nvm delete this please i got it
 
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I got the same ballpark (get it?) answer, -661J, using both common methods:

W = \int^{x_{f}}_{x_{i}} F_{k} . ds = F_{k} . s = -Fs

where F (the force vector) is given by \frac{m\Delta v}{\Delta t} where Δt is given by \Delta t = \frac{2\Delta x}{\Delta v}

as well as

W_{k} = \Delta K = -\frac{m(\Delta v)^{2}}{2}

so I would say you can most likely enter -664J assuming you minded your p's and q's properly with your significant figures. (The smallest number of significant figures I saw was actually 2, in the 68kg, so that might be an issue.)

But the work done by friction is by convention usually negative: the force of friction categorically acts in the opposite direction of the displacement, so the dot product of the friction-force function and the differential displacement will always be negative. (Because \theta = \pi.)

EDIT: I'm glad that you were able to get the correct answer on your own. That said, threads where the problem has been solved are often useful to other people who might happen upon it via Google or other means, so that they can see how to derive the solution for a similar problem. Please don't make a habit of deleting your problems and attempted solutions. And when you come up with a solution, please post it and how you got it. For example, in this case where the difference appears to have simply been the sign of your answer, that fact can be very instructive; it's often helpful for first-year physics students to see that the work done by friction has a negative value, and it's often even more helpful for them to see why it has a negative value.
 
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