Problem with energy conservation of mgh and 1/2 kx^2 (spring)

First off, you have “100N mass” which is wrong since mass is not measured in kilograms, not in Newtons. It’s force that is measured in Newtons and that force is the force of gravity I think you meant, “An object that weighs 100N is put on top of it.”

Now please excuse me while I vent a pet peeve of mine: It’s not quite correct to say “A 100 kg mass is placed on top of it.” since mass is a dynamical quantity and not a descriptive quantity. One should say instead “An object whose mass is XXX kg which therefore weighs 100N is placed on top of it.”

Now, on to bigger and better things. :)

Notice the details of how this is done. Visualize it in our mind. How are you doing this? Did you attach the weight to the unloaded spring and then let go of it? If you did then the sum of kinetic energy and potential energy of the spring and gravitational potential must be taken into account since the weight would be oscillating up and down in sinusoidal motion.

As you lower the object the gravitational force acting on the weight causes a downward force. It is this force which is compressing the spring and the force that is doing work. Since you’re lowering it at constant speed your hand must be supporting it by applying an upward force equal to the sum of the force exerted by the spring plus the force of yoru hand on the weight.

It is your hand that is doing an amount of work that is negative. Since the kinetic energy is zero when it comes to rest at the bottom then the total work done on the system is zero. The potential energy stored in the spring comes from two sources. There is energy lost by the system because the system is doing work on your hand. When all these things are taken into account you will find that energy is conserved.

Please take note that it’s the energy of a closed system that is conserved and since your hand is acting on the system the system is not closed.