# If the length of a 78 cm spring increases to 108 cm when a 6 kg weight is hanging from it, what is the spring's constant?

Jul 18, 2018

$k = 196 \text{ } \frac{N}{m}$

#### Explanation:

Given: a spring stretches from 78 cm to 108 cm when a weight of 6 kg is place on it. What is $k$?

Hooke's Law: $F = k x \text{ } \implies k = \frac{F}{x}$,

where $k$ is the spring's proportionality constant and $x$ is the distance the spring stretches.

The key to this problem is to make sure the units are correct. Force is measured in Newtons (N). $\text{ } 1 N = 1 \frac{k g \cdot m}{s} ^ 2$

This means we need units of $k g$ for the mass, and meters $\left(m\right)$ for the distance stretched.

Find distance stretched in meters:

108 cm - 78 cm = 30 cm; " "(30 cancel(cm))/1 * (1 m)/(100 cancel(cm)) = 0.3 m

Find the force exerted on the spring:

$F = \text{mass" xx "gravity} = 6 k g \times 9.8 \frac{m}{s} ^ 2 = 58.8 N$

$k = \frac{58.8 N}{0.3 m} = 196 \text{ } \frac{N}{m}$