Question

One end of a light spring of natural length $d$ and spring constant $K$ is fixed on a rigid wall and the other is fixed to a smooth ring of mass $m$ which can slide without friction in a vertical rod fixed at a distance $d$ from the wall. Initially the spring makes an angles of $37^{\circ }$ with the horizontal as shown in figure. When the system is released from rest, find the speed (in $m/s$ ) of the ring when the spring becomes horizontal.
$[\sin 37^{\circ }=3/5,\frac{g}{d}=2$ unit and $\frac{k}{m}=16$ unit]

Answer 1

Answer: 10.00

From energy conservation
At point $A$ & $B{E}_A=E_B$
$\frac{1}{2}k{x}^2+mgh=\frac{1}{2}m{v}^2$
Here $x=\ell -d=\frac{5d}{4}-d=d/4$
$h=d\tan 37=\frac{3}{4}d$
$\therefore \frac{1}{2}k\times \frac{d^2}{16}+\frac{3mgd}{2}=m{v}^2$
$v=\sqrt{\frac{k{d}^2}{16m}+\frac{3gd}{2}}$
$v=d\sqrt{\frac{k}{16m}+\frac{3g}{2d}}$