I tried to calculate the work done by friction along an arbitrary surface. and this is the result I got. This system is under the only force of gravity and friction. we can see that work done by friction is always $μmgl$ regardless of the nature of the curve so it is conservative.

so what does it mean when you say the force is conservative mathematically, I am not asking for the line integral formulation because I haven't learned it yet or the curl is zero. what I generally conclude is that(this is not related to the friction question) if the force applied is constant, i.e. same magnitude and the same direction it is definitely conservative because if it is constant it gets factored outside the integral, and $\cos\theta \,dx $ just equals $dy$ and this proves gravity is conservative.

but apparently, friction isn't conservative for some reason, where am I going wrong?

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someone help.

  • $\begingroup$ The $\mathrm ds$ that you wrote is actually the same thing as the $\mathrm d\ell$ that you wrote; the cosine is the angle that between $\mathrm ds=\mathrm d\ell$ made with horizontal, and this depends upon the path taken. So no, you have just made a mistake and not managed to prove what you tried to prove. $\endgroup$ Commented Oct 4, 2023 at 14:31
  • $\begingroup$ no ds is the infinitesimal arc length, traveled by the point and dl is the x component of that. $\endgroup$
    – Hammock
    Commented Oct 4, 2023 at 15:30
  • $\begingroup$ Why would you not choose dx for that if that is what you meant? $\endgroup$ Commented Oct 4, 2023 at 16:39
  • $\begingroup$ I don't think the variable makes a difference. dl or dx. $\endgroup$
    – Hammock
    Commented Oct 5, 2023 at 0:57
  • $\begingroup$ I find it concerning that you specifically requested the answer not to be in terms of line integrals even though that is the definition… it’s like you don’t actually want an answer. $\endgroup$ Commented Oct 5, 2023 at 2:57

2 Answers 2


As far as I understood, the path until the origin is defined once the shape of the curve is chosen. But that is only one of several possible paths if we realize that the space is 3D, and there is a z-axis perpendicular to the picture. It is not really a curve but a surface.

There are infinite other paths along the surface going from the initial point to the origin. I mean: give an impulse to the object, so that it also moves along $z$, stop it at an intermediate point Q, (with $z \neq 0$). Give another impulse from Q to the origin. The work done will change for this new path.


Friction is not conservative. The thing about a conservative force is that it changes energy back and forth between kinetic energy and potential energy. The name comes from Kinetic + potential energy being conserved.

Gravity is like this. If a mass slides without friction along a surface, gravity slows it down as it goes up. Kinetic energy becomes gravitational potential energy. As it slides back down and speeds up, the reverse occurs. You get all the kinetic energy back. Note that gravity pulls downward at all times.

A spring is like this. Electrostatic forces are like this.

Friction is not. Friction slows an object sliding along a curve to a stop. Friction is opposes the motion. If you slide the object the other way, friction switches direction and again slows the object.

Friction does not exchange energy between kinetic and potential energy. There is no such thing as "friction potential energy". Instead, friction converts kinetic energy into heat. Energy is still conserved, but not in the same way as for a conservative force.

Mathematically, there are a couple things that make a force conservative.

First, if you know the position of an object, you know the force. For gravity near the surface of the earth, it is the same everywhere so it is pretty easy to figure out. Farther away, $F \propto 1/r^2$, and points toward the Earth.

For friction, force changes direction with speed. You cannot say what the force is just by knowing where the object is.

The second condition constrains the direction of the force. If an object travels in a loop, the force might speed up the object in one part of the loop. If so, it must equally slow the object in another part. This must happen for every loop.

For gravity, any downhill part of the loop speeds the object. If the loop travels back to the start, this is balanced by the uphill part.

Mathematically, this is expressed by the curl. The curl is related the component of force along the loop. Making the curl $0$ is the same as making the sum of forces components that push the around a loop add to $0$.

Electrostatic forces are conservative. But moving charges generate magnetic fields and fields that change. These may not be conservative. Particle accelerators generate electromagnetic forces that push particles faster and faster around a loop.

  • $\begingroup$ I understand everything you wrote but many sources also state that a conservative force is a force such that the work done by it is independent of the path taken. Didn't I prove this condition for friction? is there something wrong with my work. $\endgroup$
    – Hammock
    Commented Oct 4, 2023 at 14:27
  • 1
    $\begingroup$ Push the object from A to B. For a conservative force, the work would be the same as pushing from A to B to A to B. You can see how no work around a closed loop is important. $\endgroup$
    – mmesser314
    Commented Oct 4, 2023 at 14:53
  • 1
    $\begingroup$ Also the conditions I state are enough to prove independence of path. Suppose two paths lead from A to B. Follow path 1 from A to B, and come back to A via path 2. The total work must be $0$, So the work gained on one path must be lost following the other path backward. Following it forward would gain the same as the first path. $\endgroup$
    – mmesser314
    Commented Oct 4, 2023 at 15:06
  • $\begingroup$ I can now see how both definitions are equivalent but, I still don't know how my proof is wrong. I showed that the work done by friction in that case is independent of the path taken. we obtained the expression mumgl. does his show friction is conservative $\endgroup$
    – Hammock
    Commented Oct 4, 2023 at 15:36
  • 1
    $\begingroup$ Isn't the example given by him already more than good enough? If the force is conservative, then pushing from A to B should get the same as from B to A, so total work must be 0. But if you have friction, pushing from A to B and B to A incurs double the amount of energy lost. There is no chance it will work. $\endgroup$ Commented Oct 4, 2023 at 16:40

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