Question

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?

Answer 1

`k = 196 " "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 = kx " "=> k = 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). `" "1 N = 1 (kg*m)/s^2`

This means we need units of `kg` for the mass, and meters `(m)` 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 = "mass" xx "gravity" = 6 kg xx 9.8 m/s^2 = 58.8 N`

`k = (58.8 N)/(0.3 m) = 196 " "N/m`