# Mechanical properties of Solids Concept

#### zorro

When a block of mass M is suspended by a long wire of length L, the elastic potential energy stored is given by 1/2 x Mg x l, where l is the elongation produced.
The loss in gravitational potential energy of the mass-earth system is Mgl. I wonder where does the other Mgl/2 go?

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#### Doc Al

Mentor
What do you think? Hint: How was the mass lowered from the original position to its final position?

#### zorro

The mass lowers due to its weight (stress) on the wire, which is stored as elastic potential energy of the rod.

#### Doc Al

Mentor
The mass lowers due to its weight (stress) on the wire, which is stored as elastic potential energy of the rod.
Are you just dropping the mass or are you lowering it gently?

Just dropping

#### Doc Al

Mentor
Just dropping
OK. The gravitational PE goes into both elastic PE and KE. Eventually, that KE will be 'lost' to internal energy.

#### zorro

So the elastic PE includes 2 quantities- PE due to elongation and PE due to shrinking of wire?
How is the KE lost to internal energy - You mean by the production of heat etc?

#### Doc Al

Mentor
So the elastic PE includes 2 quantities- PE due to elongation and PE due to shrinking of wire?

How is the KE lost to internal energy - You mean by the production of heat etc?
Yes. If you just drop the load, the mass will oscillate about the equilibrium point. Due to internal friction, eventually it will come to rest.

#### zorro

When the wire is stretched, there is some longitudinal elongation produced as well as decrease in cross sectional radius (shrinking in that sense).
Here the wire in not completely elastic to execute oscillations.

#### Doc Al

Mentor
You only mentioned elastic PE in your original post, so I thought that's what you were asking about. The 'additional energy' can go into various other forms, including inelastic deformation.

Thanks!

#### maimonides

Let me suggest a solution without ad hoc energy losses.
The mass m is moving under the influence of the force f = g - kx (k is the spring constant of the wire). So you can´t apply
Epot = mgh here the way you do it. If you work it out keeping this in mind, all will be ok.

Edit:
A different view of the situation: Energy has a sign (you can put energy into a system or you can take energy from it). Don´t forget you have two forces (elastic/gravity) here. Take a look at their directions.

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