# Moving along friction surfaces

If a particle moves along a one dimensional surface with constant friction. As the particle moves from point $A$ to point $B$ it loses an amount of energy equals $E(A,B)$. Consider that the particle returns back to point $A$. It loses another $E(A,B)$ units of energy so it loses total $2*E(A,B)$ of energy but doing integration to get the change of kinetic energy, We add integral of the force of friction from $A$ to $B$ to the same integral from $B$ to $A$ so we get zero. How can this happen?

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The work done against friction in moving the particle from A to B is $$E(A,B) = \int\limits_A^B F\,\mathrm{d}x = F\times(B - A).$$ It is important to note that $F$ is the force that the particle has to apply to the surface in order to overcome friction, so $F$ points in the same direction as the position vector pointing from $A$ to $B$. This ensures that E(A,B) is a positive quantity irrespective of your choice of sign conventions. The work done in moving the particle from $B$ to $A$, on the other hand, is $$E(B,A) = \int \limits_B^A (-F)\,\mathrm{d}x = \int\limits_A^B F\,\mathrm{d}x = E(A,B).$$ The total work done against friction by the particle (and hence the kinetic energy lost) is therefore $$E(A,B) + E(B,A) = 2 E(A,B) = 2 F \times (B - A).$$.