# If the length of a 42 cm spring increases to 87 cm when a 1 kg weight is hanging from it, what is the spring's constant?

Jan 10, 2016

$k = 21 , 778 N / m$

#### Explanation:

Hooke's Law states that the elongation or compression of an elastic spring is directly proportional to the resultant force acting on it and in the direction of the resultant force.
$\vec{F} = k \vec{x}$.

So in this case, the resultant force is the weight force $W = m g = 1 \times 9 , 8 = 9 , 8 N$.

The elongation in this case is $x = 87 - 42 = 45 c m$, which is $0 , 45 m$ in SI units.

Therefore by Hooke's Law, the spring stiffness constant may be given by

$k = \frac{F}{x} = \frac{9 , 8 N}{0 , 45 m} = 21 , 778 N / m$.

(This is quite a small value, indicating a very elastic spring, easy to stretch and compress. Not very stiff).