If the length of a $23 c m$ $23 cm$ spring increases to $40 c m$ $40 cm$ when a $7 k g$ $7 kg$ weight is hanging from it, what is the spring's constant?

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Answer 1 · 2018-03-14T03:30:40

$k = 4035.29$ $k = 4035.29$ kg/s^2

Hooke's Law: $F = - k x$ $F = -kx$

$F = m g$ $F = mg$ ; $m = 7$ $m = 7$ kg and $g = 9.8$ $g = 9.8$ N
$k = ?$ $k = ?$
$x = 40 - 23 = 17$ $x = 40 - 23 = 17$ cm $= 0.17$ $= 0.17$ m

The negative sign is simply for directionality. We only care about magnitude here.

$k = \frac{m g}{x} = \frac{(7) (9.8)}{0.17} = 4035.29$ $k = (mg)/(x) =((7)(9.8))/(0.17) = 4035.29$ kg/s^2

If the length of a 23cm23 cm spring increases to 40cm40 cm when a 7kg7 kg weight is hanging from it, what is the spring's constant?