Since there is no friction in the motion of this system, the total mechanical energy of the two masses will be constant ("conserved"). [Treating the pulley as "frictionless" means that it is assumed not to rotate -- which requires mechanical energy -- so the string just slides over it.]
It will be the case then that the change in kinetic and potential energy of the 2 kg. mass will equal the change in the kinetic and potential energy of the 4 kg. mass. Initially, both masses are at rest, so their starting kinetic energies (we'll call them $ \ K_2 \ $ and $ \ K_4 \ $ ) are zero. The 4 kg. mass starts at some height H above the 2 kg. mass, so we are calling the starting level of the 2 kg. mass zero. We can then call the starting potential energy of the 2 kg. mass $ \ U_2 \ = \ 0 \ $ and the starting potential energy of the 4 kg. mass $ \ U_4 \ = \ MgH \ $ , with $ \ M = 4 \ $ . This makes the initial total mechanical energy of the system
$$ K_2 \ + \ U_2 \ + \ K_4 \ + \ U_4 \ = \ 0 \ + \ 0 \ + \ 4 \cdot gH \ + \ 0 \ \ = \ 4 \cdot \ 9.81 \ H \ \ . $$
We now let the 4 kg. mass descend by 1.4 meters; this means the 2 kg. mass must rise by the same distance. Since the two masses are connected to a single string with a constant length, at every moment the speed of the two masses must be the same (the 4 kg. mass moving downward, the 2 kg. going up). So when the two masses have moved by 1.4 meters, they both have a speed $ \ V \ $ , which is what we are being asked to find. The total mechanical energy at that moment is given by
$$ K'_2 \ + \ U'_2 \ + \ K'_4 \ + \ U'_4 \ \ , $$
with
$$ K'_2 \ = \ \frac{1}{2} mV^2 \ = \ \frac{1}{2} \cdot 2 \cdot V^2 \ \ , $$
$$ K'_4 \ = \ \frac{1}{2} MV^2 \ = \ \frac{1}{2} \cdot 4 \cdot V^2 \ \ , $$
$$ U'_2 \ = \ mg \cdot 1.4 \ = \ 2 \cdot 9.81 \cdot 1.4 \ \ , $$
$$ U'_4 \ = \ Mg \cdot (H - 1.4) \ = \ 4 \cdot 9.81 \cdot (H - 1.4) \ \ . $$
The total mechanical energy at this time is equal to the total $ \ 4 \cdot 9.81 \ H \ $ we calculated for the time when the masses were released.
You'll find that the value of $ \ H \ $ we didn't specify will just cancel out (meaning it is irrelevant to the answer), and you will have an equation you can solve for $ \ V \ $ .
Try to convince yourself that the free body diagram below is correct.
- $T \to$tension of the string
- $R_1 \to$Force exerted by the pulley on the string ($4\sqrt2N$ given)
- $R \to$Normal reaction force on $P$ by the table
- $\mu R \to$Static friction working on $P$
Here $\theta=45^{\circ}$
So for the following 3 systems we've got 4 equilibrium:
\begin{align}
&\text{Pulley:}&R_1&=2T\cos\theta\tag{i}\\
&\text{Ball Q:}&T&=m_qg\tag{ii}\\
&\text{Ball P:}&T&=\mu R\tag{iii}\\
&\text{Ball P:}&R&=m_pg\tag{iv}\\
\end{align}
From $(i)$ we get :
$$4\sqrt2=2T\cos(45^{\circ}) = \frac{2T}{\sqrt2}\\
\therefore T=4N$$
From $(ii)$ we get :
$$T=m_qg=10m_q(\mathrm{taking}\ \ g=10m/s^2)\\
\therefore m_q=\frac{T}{10}=\frac{4}{10}=0.4kg$$
Best Answer
Hint.
Calling $y_P, y_Q, y_R$ and $y_O$ the heights for $P, Q, R$ and the second pulley, we have
$$ \cases{ T_1-Pg = P\ddot y_P\\ T_2-Qg = Q\ddot y_Q\\ T_2-Rg = R\ddot y_R\\ T_1 = 2T_2\\ y_P + y_O = C^{te}\\ y_Q-y_O +y_R-y_O = C^{te} } $$
NOTE
If $R$ remains at rest, then $T_2-Rg = 0$