Question

One end of a spring of natural length $2m$ and spring constant $k=100N/m$ is fixed at the ground and the other is fitted with a smooth ring of mass $1kg$ which is allowed to slide on a horizontal rod fixed at a height $2m$ (fig.) initially, then spring makes an angle of ${37}^o$ with the vertical when the system is released from rest. Find the speed (in m/s) of the ring when the spring becomes vertical.

• A. $5m/s$
• B. $3m/s$
• C. $7m/s$
• D. $6m/s$

Answer 1

The correct option is A $5m/s$

Given that,

Mass $M=1kg$

Natural length $h=2m$

Spring constant $k=100N/m$

Angle $\theta ={37}^0$

Now,

$cos\theta =\frac{h}{l}$

$l=\frac{h}{\cos \theta }$

$l=\frac{2}{0.798}$

$l=2.50$

Now, the extension

$x=l-h$

$x=2.50-2$

$x=0.5$

Now, equating KE and PE

$\frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2$

$m{v}^2=100\times {\left(0.5\right)}^2$

$m{v}^2=100\times .25$

$v^2=25$

$v=5m/s$

Hence, the speed of the ring is $5m/s$