The system shown in figure is in equilibrium. Masses m₁ and m₂ are 2 kg and 8 kg, respectively. Spring constants k₁ and k₂ are 50 N m^-1 and 70 N m^-1, respectively. If the compression in second spring is 0.5 m. What is the compression in first spring? (Both springs have natural length initially.)

As the springs have natural length initially, if one spring is compressed, the other must be expanded. Hence, the compression will be negative.

The free-body diagram of m₂ (figure) 

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \mathrm{T}+{\mathrm{F}}_2=80\mathrm{N}\text{ and }\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} {\mathrm{F}}_2=70\times 0.5=35\mathrm{N}\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{T}=80-35=45\mathrm{N}\end{array}$

FBD of m₁ (figure) 

$\begin{array}{l} \mathrm{T}+{\mathrm{F}}_1={\mathrm{m}}_1\mathrm{g}\text{ or }{\mathrm{F}}_1=-25\mathrm{N}\\ \therefore {\mathrm{X}}_1=\frac{-25}{{\mathrm{k}}_1}=\frac{-25}{50}=-0.5\mathrm{m}\end{array}$

Therefore, compression in first spring is -0.5 m. (negative sign indicates that it is extension)