Two blocks are connected by a string as shown in figure. They are released from rest. Show that after they moved a distance L, their common speed in given by sqrt((2(m_(2)-mum_(1))gl)/((m_(1)+m_(2)))) where mu is the coefficient of friction

Two blocks are connected by a string as shown in figure. They are rele

1.	Two blocks are connected by a string as shown in figure. They are released from rest. Show that after they moved a distance L, their common speed in given by sqrt((2(m_(2)-mum_(1))gl)/((m_(1)+m_(2)))) where mu is the coefficient of friction
Answer» SOLUTION :In moving a distance L by `m_(1) and m_(2)` the PE will be lost by `m_(2)` only and will be `m_(2)gl`, while work done by friction will be only on `m_(1)` and will be `MU m_(1)gL`. Now as the string is inextensible `v_(1) = v_(2) = v ` and so, the gain in KE wil be = `(1//2) (m_(1) + m_(2)) v^(2)`. By conservation of energy. Loss in PE = Grain in KE + WD against friction i.e., `m_(2)gL =(1)/(2)(m_(1)+m_(2))v^(2)+mu m_(1)gL` `v=sqrt((2(m_(2)-mum_(1))gL)/((m_(1)+m_(2))))`