If the length of a 27 cm spring increases to 45 cm when a 4 kg weight is hanging from it, what is the spring's constant?

May 21, 2017

The spring constant is $= 217.8 k g {s}^{-} 2$

Explanation:

The equation of the elongation of the spring is

$F = k \cdot \Delta x$

The force is $F = 4 g N$

The elongation is $\Delta x = 0.45 - 0.27 = 0.18 m$

The spring constant is

$k = \frac{F}{\Delta x} = \left(4 \frac{g}{0.18}\right) N {m}^{-} 1$

$= 217.8 k g {s}^{-} 2$

as

$1 N {m}^{-} 1 = 1 k g m {s}^{-} \frac{2}{m} = 1 k g {s}^{-} 2$