A ring of mass m can slide over a smooth vertical rod. The ring is connected to a spring of force constant `K=(4mg)/R` where 2R is the natural length of the spring. The other end spring is fixed to the ground at a horizontal distance 2R from the base of the rod. the mass is released at a height of 1.5 R from ground.

Options

A. work done by the spring will be `(mgR)/2`
B. work done by the spring will be `9 mgR`
C. the velocity of the ring when it reaches the ground will be `sqrt(gR)`
D. the velocity of the ring when it reaches the ground will be `2sqrt(gR)`

Answer

D

Explanation

`W_("Spring")=1/2 kx^(2)`
`=1/2xx(4 mg)/R[sqrt(4R^(2)+(9R^(2))/4)-2R]^(2)`
`=(mgR)/2`
`W_("all forces")=DeltaKE`
`(mgR)/2+mg.(3R)/2=1/2 mv^(2)`
`V=2sqrt(gR)`