Question 1002126
{{{x}}}= length of the spring when there is no load, in cm.
{{{30-x}}}= stretching of the spring with a {{{6kg}}} load, in cm.
{{{40-x}}}= stretching of the spring with a {{{10kg}}} load, in cm.
A very high load, will forever deform, or even break a spring.
However, with reasonable loads, the stretching of a spring is proportional to the load applied, applied, and vice versa.
In equation form, that proportional relation is:
Load in kg = K(strecthing in cm) ,
where K is a constant, i kg/cm.
Without writing the units, the data we are given shows
{{{6=K(30-x)}}}
{{{10=K(40-x)}}}
dividing one equation by the other, we get
{{{6/10=K(30-x)/(K(40-x))}}} ---> {{{6/10=(30-x)/(40-x)}}}
{{{6(40-x)=10(30-x)}}}
{{{240-6x=300-10x}}}
{{{10x-6x=300-240 }}}
{{{4x=60 }}} ---> {{{x=60/4}}} ---> {{{x=15}}} .
The length of the spring when there is no load is {{{highlight(15cm)}}} .