Here to solve this question we can use Work energy theorem.
Work energy theorem:- Work done by all the forces is equal to change in kinetic energy.
[tex]\begin{gathered} W(all\text{ }forces)=\text{ }change\text{ in }K.E.=\Delta K.E. \\ where,\text{ }K.E.=\text{ }Kinetic\text{ }Energy \end{gathered}[/tex]
There are only two forces acting on the sugar cube, one is gravitational force and other is friction force.
work done by gravitational force is,
[tex]W(g)=mgh[/tex]
And work done by friction is,
[tex]\begin{gathered} W(f)=Force\times Displacement\text{ }=-\mu NL \\ here,\text{ minus sign is there because the dirction of frictional force is in the opposite direction of } \\ the\text{ }displacement. \\ \mu=coefficient\text{ }of\text{ }kinetic\text{ }friction \\ N=normal\text{ }reaction\text{ }on\text{ the sugar cube}=mgcos(\theta) \\ \\ \\ \end{gathered}[/tex][tex]\begin{gathered} So,W(f)=-\mu mgcos(\theta)L \\ \end{gathered}[/tex]
Now applying work energy theorem,
[tex]\begin{gathered} W(g)+W(f)=\Delta K.E.=(K.E.)f-(K.E.)i \\ and\text{ we know that initial kinetic energy of the sugar cube = 0} \\ So,\text{ \lparen}K.E.)i=0 \\ So\text{ our new equation is} \\ W(g)+W(f)=(K.E.)f \\ So,(K.E.)f=W(g)+W(f) \\ (K.E.)f=mgh+(-\mu mLgcos(\theta)0=mgh-\mu mLgcos(\theta) \end{gathered}[/tex]
So,
Result :- Option (a) is correct option.
