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?

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Answer 1 · 2018-07-18T22:03:48

$k = 196 " "N/m$

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:

$F = "mass" xx "gravity" = 6 kg xx 9.8 m/s^2 = 58.8 N$

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

If the length of a 78 cm78 cm spring increases to 108 cm108 cm when a 6 kg6 kg weight is hanging from it, what is the spring's constant?