[LB7 (OP)]

To have the force in the other direction on the circle and on the red wall, I can rotate the circle around its center in the same time. In that case the force from friction helps to move to the right the circle. At the limit, I have no energy from friction but I have the forces, I lost d2F from the rotation of the circle and I win d1F from the translation of the circle.

[LB7 (OP)]

The method is more:

1/ draw the objects at the initial and final positions
2/ group the objects at the final position and rotate back to have the same orientation of the initial position
3/
     3.1/ place A0 the common dot, measure the distance moved by the dot of contact
     3.2/ place the center of the circle the common dot, measure the distance moved by the dot of contact
4/ calculate the algebric sum: this is the distance of friction

[LB7 (OP)]

I think the principle: the work doesn't depend of the path is false because if I decomposed the movement in 3 movement:

1/ the red arm is fixed, I move the circle along the red arm
2/ I glue the circle and the red wall, I rotate that group
3/ the red arm is fixed, I rotate the circle

in that case, the sum of energy is conserved. So the combination of 2 movement in the same time changes the result.


When I write the method:

1/ draw the start and final position
2/ group the all objects in the final position
3/ rotate the group of the final position back to have the same orientation for one object (the wall for example)
4/ rotate one object forward (the circle for example) of : the relative rotation of one object to the other less the difference of the  orientation of the dot of contact
5/ measure the distance of friction.

I need to precise, "the difference of the orientation of the dot of contact" is measured after the dot 3/, it is not the true orientation, it is the orientation after rotate back the group.