# Is friction a conservative force? [duplicate]

This question is an exact duplicate of:

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.

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

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.
• @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. – Diracology Aug 28 '17 at 17:55