Energy Transfer in a Mechanical System - Standard Pulley Scenario I understand that this is quite a basic question, I am new to dynamics and have trouble starting off questions, I found it quite difficult to find an example question alike to the one below thus I am here looking for some help.
Two masses of 2kg and 4kg are connected by a light string passing over a light frictionless pulley. If the system is released from rest determine the velocity of the 4kg mass when it has descended a distance of 1.4m.
 A: 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 \ $ .
