An elastic cord is 61 cm long when a weight of 75 N hangs from it but is 85 cm long when a weight of 210 N hangs from it. What is the "spring" constant k of this elastic cord?

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7
Mar 11, 2017

The spring constant is $= 577 N {m}^{-} 1$

Explanation:

Let the length of the elastic cord at rest $= l m$

Therefore,

In the first case, $k = {F}_{1} / \left(\Delta {x}_{1}\right) = \frac{75}{0.61 - l}$

In the second case, $k = {F}_{2} / \left(\Delta {x}_{2}\right) = \frac{210}{0.85 - l}$

We have 2 equations with 2 unknowns.

$\frac{75}{0.61 - l} = \frac{210}{0.85 - l}$

$75 \left(0.85 - l\right) = 210 \left(0.61 - l\right)$

$210 l - 75 l = 210 \cdot 0.61 - 75 \cdot 0.85$

$135 l = 64.35$

$l = \frac{64.35}{135} = 0.48$

So,

$k = \frac{75}{0.61 - 0.48} = 576.9 N {m}^{-} 1$

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