Two blocks are connected by a string, as shown in figure. They are released from rest. Show that after they have moved a distance L, their common speed is given by `v=sqrt(2(m_2-mum_1)gL//(m_1+m_2))`, in which `mu` is the coefficient of kinetic friction between the upper block and the surface. Assume that the pulley is massless and frictionless.

Explanation

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 will be `= (1//2) (m_(1)+m_(2))v^(2)`. By conservation of energy.
Loss in PE = Gain 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)-mu m_(1))gL)/((m_(1)+m_(2))))`