Question

In the situation in figure, a block of mass $1kg$ is attached to a light spring of constant $40N/m$ whose other end is fixed to the roof of a building $50cm$ 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 $\sqrt{10n}m/s$. The value of $n$ is________.

Answer 1

Answer: 3

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

$\cos \theta =\frac{OA}{OB}=\frac{0.5}{0.5+x}$ or $\frac{1}{4x}=\frac{0.5}{0.5+x}$
Or $0.5+x=2x$ or $0.5=x$
$\therefore \cos \theta =\frac{1}{4x}=\frac{1}{4\times 0.5}=\frac{1}{2}$
$\therefore \theta =60^{\circ }$
$\therefore AB=OA\tan \theta =0.5\times \sqrt{3}=\frac{\sqrt{3}}{2}$
Applying work-energy theorem,
$W=\Delta K$ or $W_s+W_F=\frac{1}{2}m{v}^2-0$
$\Rightarrow -\frac{1}{2}k(AB{)}^2+F(AB)=\frac{1}{2}m{v}^2$
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 $v=\sqrt{30}m/s$
$\therefore n=3$