Gravitational potential energy and Electric potential energy

[vishuboy]

Hi,

I've been thinking on this for a while, any help would be great :)

I have a box placed over a table. There's a pen kept over the box. What I do here is move the pen slowly, without changing its velocity and bring it down on the table.

Assuming the table and box are at a height h1 and h2 respectively. The change in potential energy is mg(h2-h1).

Here is my doubt now, there is no change in KE, but there's a energy change, that is the energy stored within the pen has decreased. My question is where does this energy remaining go?


Now coming to the electric potential energy, If a battery movies a charge q through a potential diff Va (Higher pot) to Vb (low pot). The work done is q(Vb-Va). Though there is decrease in pot energy here, the remaining energy is dissipated as thermal energy and satisfies the law of conservation of energy.

My basic doubt is, the energy which comes out as thermal energy in case of electrical circuits, like this, what happens to the remaining energy in case of table-box situation.

 

[Dale]

Because the pen was brought down slowly it did work on whatever kept it from accelerating. That work is exactly equal to the missing energy.

 

[vishuboy]

Now I get it, this means when i actually bring it down, it does work on me. To counter that my energy is expended. Am i right?

Thanks for the help :)