If the length of a 49 cm spring increases to 84 cm when a 5 kg weight is hanging from it, what is the spring's constant?

Dec 9, 2015

$140 \text{N/m}$

Explanation:

Hooke's Law states that:

$F = - k x$

$x$ is the extension which is $0.84 - 0.49 = 0.35 \text{m}$

The force $= m g = 5 \times 9.8 \text{N}$

$\therefore k = \frac{F}{x} = \frac{5 \times 9.8}{0.35} = 140 \text{N/m}$