If the length of a $23 c m$ $23 cm$ spring increases to $96 c m$ $96 cm$ when a $5 k g$ $5 kg$ weight is hanging from it, what is the spring's constant?

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Answer 1 · 2018-03-07T14:30:07

I got $67 \frac{N}{m}$ $67N/m$

We can use Hooke's Law:
$F_{e l} = - k x$ $F_(el)=-kx$
Newton's Second Law...
and the diagram:
enter image source here

At equilibrium:
$F_{e l} - W = 0$ $F_(el)-W=0$
$- k x - m g = 0$ $-kx-mg=0$
using our data and from our choice of axes (changing in meters):

$- k (- 0.73) - (5 \cdot 9.8) = 0$ $-k(-0.73)-(5*9.8)=0$

$k = \frac{5 \cdot 9.8}{0.73} = 67 \frac{N}{m}$ $k=(5*9.8)/0.73=67N/m$

If the length of a 23 cm23cm23 cm spring increases to 96 cm96cm96 cm when a 5 kg5kg5 kg weight is hanging from it, what is the spring's constant?