Is friction a conservative force?

Question

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 1

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.

Answer 2

Friction is non conservative force, because it produces path dependent work.

Let $A, B\in \Bbb R^2$. Suppose for simplicity that they are lying on $x-$axis and $O$ is between them, thus they have position vectors $\bf -a$ and $\bf a$$=(a,0),a>0,$ respectively. We choose two different paths from $A$ to $B$, along which friction is acting (against motion as usually): the line segment $\bf l$$(t)=-\bf a$$+t\bf a$, $t\in [0,1]$ and the semicircle $\bf c$$(u)=a(\cos u, \sin u)$, $u\in [π,2π]$. Then

$W(T, \bf l$$)=\int_l \bf {T} \cdot dr$$=\int_0^1 \bf T$$(\bf l$$(t))\bf l'$$(t)dt=\int_0^1 -T(t)\bf i$$\cdot\bf a$$dt=-a$$\int_0^1 T(t)dt$ and

$W(T,\bf c$$)=\int_c \bf {T} \cdot dr$$=\int_π^{2π} \bf T$$(\bf c$$(u))\bf c'$$(u)du=$ =$\int_π^{2π} -T(u)\bf i$$\cdot (-a\sin u, a\cos u)du=a\int_π^{2π} T(u)\sin udu=$ $=a\int_{-\cos π}^{-\cos 2π} Τ(-\cos u)d(-\cos u)=-a\int_{-1}^1 T(v)dv.$

Obviously $W(T, \bf l$$)\neq W(T,\bf c$$)$, hence work by friction is path dependent, so friction is non conservative force.

Answer 3

A slightly deeper way of thinking about conservative forces is to consider conservative vector fields that take the form of an exact differential,

$$Mdx + Ndy + P dz = 0 $$

Where M,N, and P are functions of x,y, and/or z. The condition for exactness is that the partial derivatives of M,N, or P w.r.t variables that they aren't coupled to are equal to the partial derivatives of M,N,or P (not the same chosen at first) w.r.t the other variables they aren't coupled to. It is easier to see written out:

$$\frac{\partial{M}}{\partial{y}}=\frac{\partial{N}}{\partial{x}}$$ $$\frac{\partial{P}}{\partial{x}}=\frac{\partial{M}}{\partial{z}}$$ $$\frac{\partial{N}}{\partial{z}}=\frac{\partial{P}}{\partial{y}}$$

With all these partial derivatives being equal it means that there is an independence of path for this vector field. If given two points A and B and the potential function (you solve for this in undergraduate differential equations) then the amount of work to get to from point A to B is always constant. Friction is different however because as some commentators have already said, " It is a path-dependent force". One of the easiest vector fields to intuitively understand is conservative gravity. In the classical limit and as long as we aren't going too high, two points above the Earth's surface always require a constant amount of work due to gravity to go from A to B. It doesn't matter if you go up 1000 feet and then come back down to point B or you go in a straight line from point A to B.

Hopefully, this helps you grow a stronger intuition for conservative forces.