What is the energy conservation principle in a two masses and pulley problem?

[vcsharp2003]

Would the 8-kg mass be a ble to lift the 14-kg mass when reaching 5 m of height on its way down?

I think if there is a large impulsive force through tension that acts on heavier mass for a very small $\triangle {t}$ time interval then yes the heavier mass would get lifted off the ground when the string becomes suddenly taut.

 

[Vigorous]

Only external forces are to be considered when applying law of conservation of energy to a system.

I got you! The work done by the tension of the rope which influence the motion of the masses m and M cancel each other. The string is inelastic, if M moves down by 5 m, m will move up by 5 m.
Energy conservation becomes:
Mgh0+mg(0)+1/2M(0)^2+1/2m(0)^2=Mgh0=Mg(0)+mgh0+1/2mv^2+1/2Mv^2 => gh0(M-m)=1/2v^2(M+m)
v=(2gh0(M-m)/(M+m))^1/2=5.169 m/s
The heavy mass hits the ground, its kinetic energy is transformed into another form of energy (heat or sound). The system will try to stay in motion at 5.169 m/s. The string connecting the two masses is no longer stretched because the weight of M is compensated by the reaction of the floor, M stays at rest. T no longer acts on m. m will uniformly decelerate with initial speed of 5.169 m/s
mgh0+1/2mv^2=mgh
h=6.36 m

 

[haruspex]

The impulsive force in this case would be a variable tension force that is going to be very high in magnitude as compared to gravity or normal reaction forces and would act for a very short interval of time just after string becomes taut. This impulse would be on each of the masses. Is this idea of impulsive force correct?

Yes.

 

[vcsharp2003]

I got you! The work done by the tension of the rope which influence the motion of the masses m and M cancel each other. The string is inelastic, if M moves down by 5 m, m will move up by 5 m.
Energy conservation becomes:
Mgh0+mg(0)+1/2M(0)^2+1/2m(0)^2=Mgh0=Mg(0)+mgh0+1/2mv^2+1/2Mv^2 => gh0(M-m)=1/2v^2(M+m)
v=(2gh0(M-m)/(M+m))^1/2=5.169 m/s
The heavy mass hits the ground, its kinetic energy is transformed into another form of energy (heat or sound). The system will try to stay in motion at 5.169 m/s. The string connecting the two masses is no longer stretched because the weight of M is compensated by the reaction of the floor, M stays at rest. T no longer acts on m. m will uniformly decelerate with initial speed of 5.169 m/s
mgh0+1/2mv^2=mgh
h=6.36 m

For part iii, final rest position will be with lighter mass at 5 m position and heavier mass on ground. This final position will be reached after multiple up and down motions of lighter and heavier masses. The impulsive force on heavier mass will lift it for each such motion, but this impulsive force will lessen after each impact of heavier mass with the ground till it's no more there. You can see that each impact of heavier mass with ground will result in dissipation of energy. At final rest position you would be left with only PE of lighter mass i.e mgh as the final total energy of the system.