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I'm having trouble calculating the work done by friction over a surface. Here is a picture I've drawn of my problem:

work-by-friction

I'm tasked with finding the work done on the box (which has a mass of 6 kg) after it goes over a surface with friction. The initial velocity is 14 m/s.

I'm aware work by friction would be equal to $Fd$, but the formula for friction is $\mu_kv$. As the net force is the friction force, I don't know how to find work by friction if the friction force varies as the box slows down. How do I find this?

Edit: Sorry, I for some reason thought the friction force was given by $\mu_kv$, when it should really be $\mu_kN$. That's probably why it wasn't making sense. I assume the work by friction would then just be $\mu_kNd$?

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closed as off-topic by Wouter, Jon Custer, Yashas, Kyle Kanos, Bill N Mar 3 '17 at 20:12

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    $\begingroup$ "The formula for friction is $\mu k(v)$". Could you elaborate on that statement? It's unusual for most of these problems to consider friction variable. Do you have the form of $k(v)$? Otherwise, you would have to write an integral... and perhaps take into account that the box will initially experience only part of the friction (before all of the box is over the region with friction, the normal force contributing to friction will be less; and you might need to consider torque). And why do you state $\mu_k=0.25$ in the diagram? That suggests it is not velocity-dependent. Please clarify. $\endgroup$ – Floris Mar 2 '17 at 22:56
  • $\begingroup$ As @Floris asks for, I would also like to hear more about the $f=\mu k(v)$ expression. Usually $\mu$ is a symbol for the friction coefficient, when friction is $f=\mu n$. And unless you change surface or load $n$, everything here is constant nomatter the speed. $\endgroup$ – Steeven Mar 3 '17 at 0:16
  • $\begingroup$ Sorry, it seems I incorrectly assumed the friction force = μk(v). I realized recently I should be using μk(Normal). Would work then be μk(Normal)(distance) ? $\endgroup$ – user6191359 Mar 3 '17 at 0:37
  • $\begingroup$ It seems to me that this question should have been closed due to insufficient effort, but to answer it: yes, the work would be $\mu_kNd$. $\endgroup$ – Wouter Mar 3 '17 at 0:54
  • $\begingroup$ Friction force does not do any work. Because it is a fixed force and cannot cause any displacement. For doing work, the force must move with the displacement due to itself. $\endgroup$ – lucas Mar 3 '17 at 10:02
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The general (scalar) version of the work formula is

$$W= \int F \; dx$$

If the force is constant, this does indeed reduce down to $W=Fx$. If not, then you have to use the integral version. In any case, you must know $F$. So, if $F$ is varying with speed, called $F(v)$, you have to know an expression/function/formula of $F(v)$ to put into the work formula.

If you are not given (or cannot find in another way) such an expression, then this question doesn't seem possible to solve.

If on the other hand the force is just usual friction, which is constant regardless of e.g. speed, then we can quickly find the work without the integral as:

$$W=Fx=\mu n x$$

Where $n$ is the normal force (which you must find with a separate calculation).

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If the force is indeed dependent on velocity then you cannot calculate the work unless you know the explicit dependence F(v). However, in many textbook problems the friction force is considered independent of v. What is the context of your problem? Where does it come from?

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  • $\begingroup$ This feels more like a comment than an answer to the question. $\endgroup$ – Floris Mar 2 '17 at 22:53
  • $\begingroup$ It says pretty much the same as the answer above. Does it need to have some minimum word count to be an "answer"? $\endgroup$ – nasu Mar 3 '17 at 14:11
  • $\begingroup$ There is no minimum word count for a "good" answer, but it is clear from the fact that the OP accepted the other answer that it was more helpful. The "system" is far from perfect but I recommend that you look around and see what works for other people. I read your answer, and gave you some advice. You are free to ignore it. $\endgroup$ – Floris Mar 3 '17 at 14:30

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