Work Energy And Power Question 267
Question: A ring of mass m can slide over a smooth vertical rod as shown in figure. The ring is connected to a spring of force constant $ k=4mgIR $ , where 2 R is the natural length of the spring. The other end of spring is fixed to the ground at a horizontal distance 2 R from the base of the rod. If the mass is released at a height 1.5 R, then the velocity of the ring as it reaches the ground is Critical Thinking
Options:
A) $ \sqrt{gR} $
B) $ 2\sqrt{gR} $
C) $ \sqrt{2gR} $
D) $ \sqrt{3gR} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Initial length of the spring, $ {\ell _{i}},=\sqrt{{{(2R)}^{2}}+{{(1.5R)}^{2}}} =2.5R $
$ \therefore x _{i}=2.5R-2R=0.5R. $
$ Now\frac{1}{2}kx_i^{2}+mg( 1.5R )=\frac{1}{2}mv^{2} $ or $ \frac{1}{2}k{{( 0.5R )}^{2}}+mg( 1.5R ) $
BETA
