Physics - NLM Question with Solution | TestHub

Now, let the length of the spring in the given position be $\ell$. From the figure:

$\ell \cos 53^{\circ }=\ell \Rightarrow \ell =5/3m$

Elongation in the spring: $5/3-1=2/3m$

So, the spring force: $kx=30\times 2/3=20N$

Free Body Diagram

(FBD) of the block:

$$\begin{array}{rl}20\cos 53^{\circ }+N& =2g\\ N& =20-20\cos 53^{\circ }\\ N& =8N\\ 20\sin 53^{\circ }-f& =ma\\ 20\sin 53^{\circ }-\mu N& =ma\\ 20\times 4/5-1\times 8& =2\times a\\ \Rightarrow a& =4m/s^2\end{array}$$