Sunday, November 20, 2016

energy conservation - Is friction a conservative force?



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 $x$.



Answer



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.


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