Answer
(a) $d=\frac{(m_1+m_2)v^2}{2m_2g}$
(b) positive
(c) $\frac{1}{2}m_1v^2$
Work Step by Step
(a) We know that
$K.E_i+U_i=K.E_f+U_f$
$\implies \frac{1}{2}(m_1+m_2)v^2+0=0+m_2gd$
This simplifies to:
$d=\frac{(m_1+m_2)v^2}{2m_2g}$
(b) We know that the rope exerts a force on $m_2$ in the same direction in which it moves. Hence the work done by rope is positive.
(c) We can find the required work done as
$W_{nc}=m_2gd-\frac{1}{2}m_2v^2$
$\implies W_{nc}=\frac{1}{2}(m_1+m_2)v^2-\frac{1}{2}m_2v^2$
$\implies W_{nc}=\frac{1}{2}m_1v^2$
