# Homework Help: Spring constant

1. Nov 28, 2008

### Naldo6

An elastic cord is 67.7 cm long when a weight of 58.7 N hangs from it and is 85.8 cm long when a weight of 88.1 N hangs froms it. What is te spring constant of this elastic cord?

how i know which is the long of the spring when it is in equilibrium to then calculate the sprin constant?...

or anyother help to solve this problem

2. Nov 29, 2008

### horatio89

Hello once again,

A review of Hooke's Law will reveal that the k, the spring constant, is a constant of proportionality. This means that F_1/x_1 = F_2/x_2 where x is the extension of the spring. Does this shed any light?

Regards,
Horatio

3. Nov 29, 2008

### Naldo6

ok but if i do that what i can get:

because

58.7 N / .677 m = 88.1 N / .858 m

and that doesn't help me too much... because i dont know the distance of the spring when it is at equilibrium

4. Nov 29, 2008

### Naldo6

how i know what is the extension of the spring?...

5. Nov 29, 2008

### horatio89

That's not exactly true. Remember, x is the extension of the spring, not the length of the spring. So, if we let the equilibrium value be l , the extension in one case is (0.677 - l). Does that clear things up?

6. Nov 29, 2008

### Naldo6

ok i know that x is the value of the extension of the spring but how i know whta is the extension of the spirng when is at equilibrium to then calculate the extension from the diferencies of my value sof x with the equilibrium distance?....

7. Nov 29, 2008

### horatio89

Since the extension x = (length of spring with load - length of spring at equil.):

The eqn shld be:

58.7 / (0.677 - l) = 88.1/ (0.858 - l)

where l is the equil. length, solve for l, with is simple algebraic manipulation...

Last edited: Nov 29, 2008
8. Nov 29, 2008

### horatio89

edited the eqn, sorry, made a typo

9. Nov 29, 2008

### Naldo6

ok let me try and tell u....

10. Nov 29, 2008

### Naldo6

solving for l y get 0.316.... then i use F=kx solve for k and get

k= F/x = 58.7/(0.677-0.316)=162.60 is this right?

11. Nov 29, 2008

### horatio89

12. Nov 29, 2008

### Naldo6

ok many thanks....