In the figure below, find the velocity of m₁ in m/s when m₂ falls by 9 m.

Using virtual work method, 

${\mathrm{Ta}}_1-4{\mathrm{Ta}}_2=0$

$$\begin{gathered} \Rightarrow {\mathrm{a}}_1=4{\mathrm{a}}_2 \\ \mathrm{or} {\mathrm{s}}_1=4{\mathrm{s}}_2=4\times 9=36\mathrm{m} \end{gathered}$$

Applying second law of motion gives,

$$\begin{gathered} \mathrm{T}-\mathrm{\mu mg}={\mathrm{ma}}_1 \\ \mathrm{and} \\ 2\mathrm{mg}-4\mathrm{T}=2{\mathrm{ma}}_2 \end{gathered}$$

After solving we get, ${\mathrm{a}}_1=\frac{32}{9}\mathrm{m}/{\mathrm{s}}^2$

Using equation of Kinematics, 

$$\begin{gathered} {\mathrm{v}}^2={\mathrm{u}}^2+2{\mathrm{a}}_1{\mathrm{s}}_1 \\ \Rightarrow \mathrm{v}=16 \mathrm{m}/\mathrm{s} \end{gathered}$$

Alternate solution, 

$\frac{1}{2}{\mathrm{m}}_2{\mathrm{v}}^2+\frac{1}{2}{\mathrm{m}}_1(4\mathrm{v}{)}^2={\mathrm{m}}_2\mathrm{g}\times \mathrm{y}-{\mathrm{\mu m}}_1\mathrm{g}(4\mathrm{y})$

$\text{ where }\mathrm{y}=9\mathrm{m}$

$\Rightarrow \mathrm{v}=16{\mathrm{ms}}^{-1}$