I don't understand this passage from the book I'm reading:

Physically, it is clear that a system cannot be conservative if friction or other dissipation forces are present, because $\mathbf{F}\cdot d\mathbf{s}$ due to friction is always positive and the integral cannot vanish.

Why Is the work done by friction always positive? If $\mathbf{F}$ is opposite to the force moving the body, the sign should vary depending on the sign of $d\mathbf{s}$

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    $\begingroup$ The work done by friction relative to the surface the friction is acting on can't ever be positive. It's probably a typo. $\endgroup$ Commented Apr 3, 2019 at 11:45
  • $\begingroup$ ${}$ Which book? $\endgroup$
    – Qmechanic
    Commented Apr 3, 2019 at 12:45

1 Answer 1


The force referred to by the book is probably an external force applied to the object and not the friction force. When the external force equals the kinetic friction force it pushes the object at constant velocity. It does positive work on the object equal to Fs. But the surface does an equal amount of negative work on the object per @Aaron Stevens which is converted to heat, so the net work done on the object is zero.

This is also explained by the work energy theorem. The net work done on an object equals its change in kinetic energy. Since the velocity of the object is constant the change in kinetic energy is zero and therefore the net work done on the object is zero.

So my guess is the book may be referring to the work done by an external force on the object against friction, and not the work done by friction.

By the way I found another post quoting Goldstein’s book on classical mechanics where he talked about work against friction being positive. He was referring to an external force.

Hope this helps.

  • $\begingroup$ @AaronStevens What if I was to rephrase it to say I do positive work on the object transferring energy to it while at the same time the surface does the same magnitude of negative work on the object transferring the energy from the object to the surface in the form of heat. $\endgroup$
    – Bob D
    Commented Apr 3, 2019 at 18:31

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