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Suppose a block of mass $m$ is being pulled on a hill by a force $F_{app}$, the block is being pulled slowly such that $\Delta KE = 0$.

Our teacher showed that the expression of work done by friction $W_{fric}$ is independent of the path traversed by the block, which is not a characteristic of non conservative force. How is this possible?

Working: $$ \Delta KE = 0 $$ so,$$W_n + W_g + W_{fric}+ W_{app}= \Delta KE = 0 $$(from work-energy theorem). As $$W_n=0$$(work done by normal Force), $$0+W_g +W_{fric} + W_{app} =0\, .$$ Now, $W_g$ is independent of path and is equal to $-mgh$ so, $$W_{app} = -(mgh + W_{fric})\, .$$ Now for $W_{fric}$, $N=mg \cos(\theta)$ where $\theta$ is the inclination of hill slope wrt positive $x$-axis, $$ dW_{fric} = kmg\cos(\theta) cos(180) ds\, , $$ (from $W = fs\cos(\theta)$ and $F_{fric} = kN$.) $ds$ is small displacement along the slope, so $ ds \cos(\theta) $ is small displacement along positive $x$-axis, Reordering our last equation, $$ dW_{fric} = -kmg dx$$ as $\cos(180) = -1$ and $ds \cos(\theta) = dx$ Integrating on both sides, $$\int{dW_{fric}} = -kmg \int dx$$ so $W_{fric} = -kmgx$ which does not depend on the length of path taken but only on the horizontal displacement $x$.


marked as duplicate by Qmechanic Aug 28 '17 at 17:29

This question was marked as an exact duplicate of an existing question.

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    $\begingroup$ $ds\cos{\theta} = d \theta$, How is this possible ? Please review your penultimate equation. It should be $dx$ in the final integral. $\endgroup$ – Mitchell Aug 28 '17 at 17:17
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    $\begingroup$ Is this a rewrite of an earlier version, just curious thanks? $\endgroup$ – user167453 Aug 28 '17 at 17:19
  • $\begingroup$ @Farcher has an answer to the previous incarnation of this question: physics.stackexchange.com/a/354129/36194 $\endgroup$ – ZeroTheHero Aug 28 '17 at 17:31
  • $\begingroup$ I saw that, he says correct but how can you defy the above derivation? $\endgroup$ – drake01 Aug 28 '17 at 17:33
  • $\begingroup$ @Mitchell edited $\endgroup$ – drake01 Aug 28 '17 at 17:34

Either you misunderstood your teacher or he made a mistake. Work done by friction is path dependent. That is why friction is non conservative.

In your example, consider two trajectories from $A$ to a point $B$ immediately above. One trajectory goes straight from $A$ to $B$ and the work due to friction is a small negative amount. For the second trajectory consider a path starting from $A$, going horizontally far and far away from $A$, going uphill and then returning horizontally to $B$. The work due to friction would be huge negative amount.

In general, the work done by a force is path independent if and only if the work done on any closed curve vanishes. Note that friction is always opposite to the motion so its work will be negative for any curve, in particular, for any closed curve.

  • $\begingroup$ I understand your point of view, but according to the derivation in the question, the contrary also seems to be true at least in the case given in question, is it a special case? $\endgroup$ – drake01 Aug 28 '17 at 17:36
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    $\begingroup$ @user167650 Your mistake is not to be careful with line integrals. Since friction is opposite to the velocity, write it as $\vec F=-F\hat v$. The arbitrary displacement can be written as $d\vec s=ds\hat v$. Thus $dW=\vec F\cdot d\vec s=-F ds$ and $W=-F\int ds=-Fl$ where $l$ is the length of the curve. $\endgroup$ – Diracology Aug 28 '17 at 17:55

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