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A ring of mass m can slide over a smoot...

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 = 4 mg//R`, where `2R` is the natural length of the spring . The other end of spring is fixed to the ground at a horizontal distance `2R` from base of the rod . If the mass is released at a height `1.5 J` then the velocity of the ring as it reaches the ground is

A

`2sqrt(gR)`

B

`sqrt(gR)`

C

`(sqrt(gR))/(2)`

D

`(3sqrt(gR))/(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Let v be velocity of the ring as it reaches the ground. From work energy theorem,
`W_(gravity)+W_(spring)=triangleK`
`impliesmg((3R)/(2))+(1)/(2)kx^2=(1)/(2)mv^2`
`implies(3mgR)/(2)+(mgR)/(2)=(1)/(2)mv^2`
`impliesv=2sqrt(gR)`
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