91Ó°ÊÓ

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 ${\stackrel{→}{F}}_s$and the gravitational force ${\stackrel{→}{F}}_g$.

Let us find the angle $\mathrm{θ}$which is shown in the free body diagram from the left triangle shown in the figure.

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

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

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

Therefore, the net force is

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

$F_T\cos \mathrm{θ}=mg$ ….. (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

${\stackrel{→}{F}}_T=kx$

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

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

$$\begin{gathered} {\stackrel{→}{F}}_T\cos \mathrm{θ}=mg \\ kx\cos \mathrm{θ}=mg \end{gathered}$$

$x=\frac{mg}{k\cos \mathrm{θ}}$

Now you can plug our values for $m,g,k$and $\mathrm{θ}$into the above equation.

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

$x=4.7\text{cm}$

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 $10.3\text{cm}$.