A mass less spring when stretched by 150 N shows an extension of 1.5 m. This spring is arranged as shown in figure. A block of mass 10 kg is released at rest on friction less incline.  It slides down and compresses the spring by 3 m before coming to rest momentarily, distance by which the block slides before hitting the spring is $(takeg=10m/s^2)$ ___

$Given stretching force =150N$

$extension in spring is x=1.5m$

$force in spring = Kx$

$$\begin{gathered} 150=K(1.5) \\ \Rightarrow K=100 N/m \end{gathered}$$

$$\begin{gathered} KE of mass is stored as PE of spring \\ \Rightarrow \frac{1}{2}m{V}^2=\frac{1}{2}K{x}^2 \end{gathered}$$
$\Rightarrow V^2=\hspace{0.17em}\frac{K{x}^2}{m}$

$$\begin{gathered} given x=3 m; mass m=10 kg \\ \Rightarrow V^2=\hspace{0.17em}\frac{100{\left(3\right)}^2}{10} \\ \Rightarrow V^2=90 \end{gathered}$$
$acceleration of block is a=g\sin \theta$
$equation of motion is V^2=2aS$
               
$\Rightarrow S=\frac{V^2}{2a}$

$S=\frac{90}{2g\sin \theta }$

$S=\frac{90}{2\times 10\times \sin 30°}$

$\mathrm{S}=9\mathrm{m}=\mathrm{distance} \mathrm{by} \mathrm{which} \mathrm{block} \mathrm{slides}$