# 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 $x$.

• Is this a rewrite of an earlier version, just curious thanks?
– user167453
Commented Aug 28, 2017 at 17:19
• @Farcher has an answer to the previous incarnation of this question: physics.stackexchange.com/a/354129/36194 Commented Aug 28, 2017 at 17:31
• First of all the work done by gravity in this case is $mg \sin \theta s$ where $s$ is the distance along the hill. To determine whether a force is conservative or not you must show that it does not depend on the details of any possible path between two endpoints. What you have done is calculated the equation for work done by friction along a certain path i.e your work does not prove your point Commented Aug 28, 2017 at 17:54
• It proves that work done by friction depends only on the horizontal displacement not on the inner details of the path Commented Aug 28, 2017 at 18:33

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.

• 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? Commented Aug 28, 2017 at 17:36
• @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. Commented Aug 28, 2017 at 17:55

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

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

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.