Calculate Spring Constant: 58.7N & 88.1N Weights

AI Thread Summary
To calculate the spring constant of an elastic cord under two different weights, the equilibrium length of the spring must be determined. Using Hooke's Law, the relationship between force and extension is established as F1/x1 = F2/x2. The extension is calculated by subtracting the equilibrium length from the lengths of the spring under load. After solving the equations, the equilibrium length is found to be 0.316 m, leading to a spring constant (k) of approximately 162.60 N/m. The calculations confirm the accuracy of this result.
Naldo6
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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
 
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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
 
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 don't know the distance of the spring when it is at equilibrium
 
how i know what is the extension of the spring?...
 
Naldo6 said:
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 don't know the distance of the spring when it is at equilibrium

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?
 
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?...
 
Naldo6 said:
58.7 N / .677 m = 88.1 N / .858 m

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:
edited the eqn, sorry, made a typo
 
ok let me try and tell u...
 
  • #10
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
Yes, your answer agrees with my calculated answer.
 
  • #12
ok many thanks...
 
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