In the system shown in the figure the mass $m$ moves in a circular arc of angular amplitude $60°$. Mass $4 \mathrm{m}$ is stationary . Then:

,$$\begin{gathered} 1) Conserving energy ,\frac{1}{2}m{v}^2=mgl(1-\cos {60}^0) \Rightarrow \frac{m{v}^2}{l}=mg \\ Therefore when the mass is at B , \mu .4mg = mg +\frac{m{v}^2}{l} \Rightarrow \mu =0.5 \end{gathered}$$

$2)$The work dine by gravitational force in the block $\mathrm{m}$ is positive when it moves from $A$ to $B$ as the angle between  gravitational force and displacement is acute.

$3)$The power delivered by the tension is P =$\overrightarrow{T}.\overrightarrow{v}.$ At position  $B$, $\overrightarrow{T} and \overrightarrow{v} are mutually perpendicular .Hence P$ is zero.

$4)$Power delivered by T is always zero .Therefore by work energy theorem ,the kinetic energy of $\mathrm{m}$ in position $B$ equals the work done by gravitational force on the block when its moves from position $A$ to $B$.