# Work done by force on block against gravity and friction up irregular rough incline?

I recently found a question which I think is from Irodov which asked about the work done by a force $F$ on a body of mass $M$ up a rough hill. At each point in the path of the body the force is tangential to to the hill. Now in the solution to this question, it is eventually proven that the work done by the force $F$ is $\mu mgx + mgy$, hence proving that the path trajectory does not matter in the net work done.

The work done against friction is found to be $\mu mgx$ through integration of the work done against friction $\mu mg\cosθ\,{\rm d}l$ where ${\rm d}l$ is a minute distance travelled by the body and $\mu mg\cosθ$ gives the normal force and I understood the method.

However all this while I have learnt that work done by a non-conservative force such as friction depends on the path traversed. Is that incorrect? Or if it is correct and applied only to certain cases, what cases is it applied to? What has gone wrong in my understanding?

• Work done by friction is of course path dependent. ... The term $x$ in the work done by friction indicates so ... May 8, 2018 at 13:48
• In this example only a motion straight up the slope is to be considered. May 8, 2018 at 14:09
• The work done here just depends on the distance travelled in the x direction, but it does not matter how bumpy the incline is. However in many books it says that while conservative forces like gravity are path independent, friction depends on the route taken. So I am confused.
– Hema
May 8, 2018 at 14:10
• @Farcher motion straight up the slope meaning?
– Hema
May 8, 2018 at 14:11
• What do you mean by $x$ and $y$ ? May 8, 2018 at 14:16