Find net work done

Net work done = change in potential energy + change in kinetic energy + work done by friction

In case of minimum work done , it means that the particle is not transferred any more energy than needed or that it is slowly moved with practically 0 kinetic energy or speed at all instants

hence change in kinetic energy = 0-0 =0

Now since A and B are at the same height

change in gravitational potential energy = 0

Now work done by friction = net work done

Now consider any arbitary route with slope angle at any particular point as

$\theta '\\ \\ Then\quad the\quad frictional\quad force\quad acting\quad on\quad it\quad is\quad \\ \\ \mu mg\quad cos\quad \theta \quad \\ \\ and\quad the\quad work\quad done\quad in\quad an\quad elementary\quad displacement\quad 'dl'\\ along\quad track\quad is\quad \mu mg\quad cos\quad \theta \quad dl\quad \\ =\quad \mu mg\quad (dl\quad cos\quad \theta )\quad =\quad \mu mg\quad dx\\ integrating\quad we\quad get\quad \\ \\ \quad \mu mg\quad (x2\quad -x1)\quad which\quad in\quad this\quad case\quad is\quad \mu mg\quad AB\quad =\quad (0.2)(10)(2)\quad =\quad 4J$

Its amazing how work done by friction is independent of path length and only depends on displacement ,, just like the force on any current carrying conductor in magnetic field is same as long it joins the same points no matter how it does

Hello Saketh,I don't think the answer is 4. I got it as 2.76.
We don't have to do work on moving through the hemispheres. so net distance moved is2-(.4+.2+.02)=2-0.62=1.38 WD =.2 x1 x 10x1.38=2.76