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I`m trying to write an expression ( Work & Energy ) of spring that is not loose and it has a box on it with a weight of 1Kg.
how its should be? this is what I tried to do:
$$\frac{kx^2}{2}-mg$$ this is an illustration of it.
enter image description here

the $m$ box is $1kg$ and $k=200$ and the state is that the spring is not loose. I would like to get some advice how to do that.
thanks!


EDIT
Here is my full problem. I want to find the maximum contraction of the spring, The first mass falls on the mass already in the spring. so I chose to solve it with Work and Energy.
$$mg(3+x_{0})+\frac{k{x_{0}}^2}{2}-mg=\frac{k{x_{0}}^2}{2}-3g$$ this is the right expression to write?
E$_{end}$=E$_{start}$?


the full problem

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may you define $k$? –  al-Hwarizmi Jun 15 '13 at 11:49
    
Elastic coefficient –  Ofir Attia Jun 15 '13 at 11:53
1  
and what is $kx^2/2$ and its dimension? –  al-Hwarizmi Jun 15 '13 at 11:55
    
Work of spring $ W = \int^t_0 Fvdt = \int^t_0 kxv_{x}dt=\frac{1}{2}kx^2$ en.wikipedia.org/wiki/Work_(physics) –  Ofir Attia Jun 15 '13 at 12:02
    
@OfirAttia : Don't mix variables and numerical values: it is unreadable : When we see an expression, we must check directly if the units are correct, in your expressions, we see together $mg(3+x0), mg, 3g$. This is a nonsense. If you use variables, use variables only. If, at the end, you want to test numerical values, use numerical values for everything, but do not mix the 2. –  Trimok Jun 15 '13 at 17:44

1 Answer 1

$$W = \int^t_0 Fvdt = \int^t_0 kxv_{x}dt=\frac{1}{2}kx^2$$ $$G=mg$$

Then under static balance:

$W+G=0$

hence:

You are correct!

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ok, if its right to write it like that, I added my full problem in Work and Energy, if you can write me some comments on that. thanks. –  Ofir Attia Jun 15 '13 at 12:21
    
what is your $G$ and your $W$ in the new balance situation? –  al-Hwarizmi Jun 15 '13 at 13:08
    
@al-Hwarizmi : Same remark as for Ofir Attia (see above) : Don't mix variables and numerical values: it is unreadable. (mg has not the dimension of an energy) –  Trimok Jun 15 '13 at 17:46
    
I am myself meanwhile confused. The set up of the question needs to be really calibrated to a norm of notation. Sorry but cannot help more. –  al-Hwarizmi Jun 15 '13 at 18:20

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