Question

In the situation in figure, a block of mass $1\, kg$ is attached to a light spring of constant $40 \,N / m$ whose other end is fixed to the roof of a building $50 \,cm$ above the smooth horizontal surface. Initially, spring is in natural length and vertical. When a force $F=20 \sqrt{3} N$ is applied on the block, the block starts to move. The speed at the instant it breaks off the surface below it is $\sqrt{10 n} \,m / s$. The value of $n$ is________.

Answer 1

If the block leaves the surface at point $B$, then
$\therefore k x \cos \theta=m g (\because N=0)$
Or $40 x \cos \theta=10$
Or $x \cos \theta=\frac{1}{4}$
From the figure,

$ \cos \theta=\frac{O A}{O B}=\frac{0.5}{0.5+x} $ or $\frac{1}{4 x}=\frac{0.5}{0.5+x}$
Or $0.5+x=2 x $ or $ 0.5=x $
$\therefore \cos \theta=\frac{1}{4 x}=\frac{1}{4 \times 0.5}=\frac{1}{2}$
$\therefore \theta=60^{\circ} $
$\therefore A B=O A \tan \theta=0.5 \times \sqrt{3}=\frac{\sqrt{3}}{2}$
Applying work-energy theorem,
$W=\Delta K $ or $W_{s}+W_{F}=\frac{1}{2} m v^{2}-0$
$\Rightarrow -\frac{1}{2} k(A B)^{2}+F(A B)=\frac{1}{2} m v^{2}$
Or $-\frac{1}{2} \times 40 \times \frac{3}{4}+20 \sqrt{3} \times \frac{\sqrt{3}}{2}=\frac{1}{2} \times 1 \times v^{2}$
Or $-15+30=\frac{v^{2}}{2}$ or $v=\sqrt{30} \,m / s$
$\therefore n=3$