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?

1 Answer
Jul 18, 2018

Answer:

#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#