Question

A small bar A resting on a smooth horizontal plane is attached by thread to a point P and by means of a weightless pulley, to a weight $B$ possessing the same mass as the bar itself. Besides, the bar is also attached to a point 0 by means of a light non-deformed spring of length $l_0=50cm$ and stiffness $k=5mg/l_0,$ where $m$ is the mass of the bar. The thread PA have been burned, the bar starts moving. Find it's velocity at the moment when it is breaking off the plane.

• A. $1.2m/s$
• B. $0.4m/s$
• C. $0.8m/s$
• D. $1.7m/s$

Answer 1

The correct option is D $1.7m/s$


Equating verical force on block,
$kx\sin \theta =mg$
$\begin{array}{cc}x& =\frac{mg}{k\sin \theta }\\ & =\frac{mg\times l_0}{5mgsin\theta }\end{array}$
$=\frac{l_0}{5\sin \theta }$

$\sin \theta =\frac{l_0}{l_0+x}$

$x=\frac{l_0}{5\left(l_0\right)}(l_0+x)$

$\Rightarrow 5x=l_0+x$
$\Rightarrow 4x=l_0$
$\Rightarrow x=\frac{l_0}{4}$

Conserving energy of the system at this position,
$0=\frac{1}{2}k{x}^2+2\left(\frac{1}{2}m{v}^2\right)-mgd$

$\Rightarrow \frac{1}{2}\frac{5mg}{l_0}\times {\left(\frac{l_0}{4}\right)}^2+m{v}^2-mgd=0$

$\Rightarrow \frac{5}{2}\frac{mg}{l_0}\times \frac{l_0^2}{16}+m{v}^2-mgd=0$

$\Rightarrow \frac{5}{32}mg{l}_0+m{v}^2-mgd=0$

Also
$d=\sqrt{{(l_0+x)}^2-l_0}$

$=\sqrt{x^2+2l_{0}x}$

$=\sqrt{(\frac{l_0}{4}{)}^2+2l_0(\frac{l_0}{4})}$

$=\frac{3l_0}{4}$

Now
$\frac{5}{32}mg{l}_0+m{v}^2-mg\frac{3l_0}{4}=0$

$\begin{array}{cc}\Rightarrow v^2& =\frac{3}{4}{l}_{0}g-\frac{5}{32}{l}_{0}g\\ & =l_{0}g(\frac{3}{4}-\frac{5}{32})\end{array}$

$⟹v^2=\frac{1}{2}\times 10\times \left(\frac{19}{32}\right)$

$⟹v=1.7m/s$

For detailed solution watch the next video.