Where does the missing energy go?

[audjobman]

Where does the missing energy go?!

When suspending a mass from a spring, we can determine the distance the spring is stretched by using mg=kx. However, if we consider the energy transferred, the gravitational potential energy lost by the mass should equal the energy stored in the spring, no?

But when I use \DeltaE_g=mg\Deltah and E_p=½2x2, and I consider that the change in height for the mass and the distance the spring is stretched are the same, then:

mgx=½kx^2

Simplifying this gives me mg=½kx. So, where does the other half of the energy go?

I've tried considering that it must be converted to kinetic energy or thermal energy but I've conducted the experiment and, when the mass is gently lowered to rest, there is no significant heating of the spring that I can see and there is no apparent kinetic energy. Am I missing something?

I'm certain I am. Please help me find my way...

 

[Vanadium 50]

The force increases with distance. The average force is not kx but half that.

 

[Doc Al]

I've tried considering that it must be converted to kinetic energy or thermal energy but I've conducted the experiment and, when the mass is gently lowered to rest, there is no significant heating of the spring that I can see and there is no apparent kinetic energy. Am I missing something?

When you gently lower the mass, you exert a force on the mass which does negative work. You essentially absorb that energy in your muscles, turning it into internal energy ("heat"). If you didn't exert that force--if you just let the mass drop onto the spring--then the mass would end up with kinetic energy.

 

[Bob_for_short]

When suspending a mass from a spring, we can determine the distance the spring is stretched by using mg=kx. However, if we consider the energy transferred, the gravitational potential energy lost by the mass should equal the energy stored in the spring, no?

But when I use \DeltaE_g=mg\Deltah and E_p=½2x2, and I consider that the change in height for the mass and the distance the spring is stretched are the same, then:

mgx=½kx^2

Simplifying this gives me mg=½kx. So, where does the other half of the energy go?

I've tried considering that it must be converted to kinetic energy or thermal energy but I've conducted the experiment and, when the mass is gently lowered to rest, there is no significant heating of the spring that I can see and there is no apparent kinetic energy. Am I missing something?

I'm certain I am. Please help me find my way...

The equation mg=kx gives you the equilibrium distance, let us call it x₀. Of course, the potential energy at this position is mgx₀=(1/2)kx₀².

But if you attach the mass at x=0 and let it go, the force mg will make larger work: the mass will pass by the equilibrium position and will go down, to 2x₀. Then the mass will oscillate around the equilibrium position. While passing x₀, the mass will also have a kinetic energy.

Bob.

 

[Dadface]

And extending Doc als and Bob for shorts answers the vibrations will die down as energy is lost to the surroundings and the mass will eventually come to rest at its equilibrium position.