A ball of mass 450 g hangs from a spring whose stiffness is $\mathbf{110}\mathbf{}\mathit{N}\mathbf{/}\mathit{m}$. A string is attached to the ball and you are pulling the string to the right, so that the ball hangs motionless, as shown in Figure. In this situation the spring is stretched, and its length is15 cm. What would be the relaxed length of the spring, if it were detached from the ball and laid on a table?

Mass, $m=450\text{g}=0.45\text{kg}$.

Spring of stiffness, $K=110N/m$.

Draw the free body diagram as below.

The system is the ball and the surroundings are the tension force of the spring $F_T$, the pulled force by the string ${\overrightarrow{F}}_s$and the gravitational force ${\overrightarrow{F}}_g$.

Let us find the angle $\theta$which is shown in the free body diagram from the left triangle shown in the figure.

$$\begin{gathered} \sin \theta =\frac{8\text{cm}}{15\text{cm}} \\ ={\sin }^{-1}\left(0.533\right) \\ =32.2° \end{gathered}$$

The tension force $F_T$has two components ${\overrightarrow{F}}_{T,x}and{\overrightarrow{F}}_{T,y}$. The vertical forces in the x-direction are equal to each other according to the momentum principle. Therefore

$$\begin{gathered} \frac{d\overrightarrow{p}}{dt}=F_{net,y} \\ =0 \end{gathered}$$

Therefore, the net force is

$$\begin{gathered} {\overrightarrow{F}}_{net,y}={\overrightarrow{F}}_{T,y}-{\overrightarrow{F}}_g \\ {\overrightarrow{F}}_{T,y}={\overrightarrow{F}}_g \end{gathered}$$

$F_T\cos \theta =mg$ ….. (1)

Here g is the acceleration due to gravity having a value $9.8m/s^2$.

The tension force $F_T$in the spring is given by

${\overrightarrow{F}}_T=kx$

Here k is the stiffness of the spring and x is the stretched length in the spring.

Plug the above expression of ${\overrightarrow{F}}_T$into equation (1) to get x.

$$\begin{gathered} {\overrightarrow{F}}_T\cos \theta =mg \\ kx\cos \theta =mg \end{gathered}$$

$x=\frac{mg}{k\cos \theta }$

Now you can plug our values for $m,g,k$and $\theta$into the above equation.

$$\begin{gathered} x=\frac{mg}{k\cos \theta } \\ =\frac{\left(0.450\text{kg}\right)\left(9.8m/s^2\right)}{\left(110N/m\right)\left(\cos 32.2°\right)} \\ =\frac{4.41N}{\left(110N/m\right)0.846} \\ =0.047m \end{gathered}$$

$x=4.7\text{cm}$

The relaxation length of the spring is equal to the original length l minus the stretched length x. Therefore

$$\begin{gathered} R=l-x \\ =15\text{cm}-4.7\text{cm} \\ =10.3\text{cm} \end{gathered}$$

Hence, the relaxed length of the spring is $10.3\text{cm}$.