On end of a light spring of natural length d and spring constant k is fixed on a rigid wall and the other is attached to a smooth ring of mass m which can slide without friction on a vertical rod fixed at a distance d from the wall. Initially the spring makes an angle of `37^@` with the horizontal. as shown in figure. When the system is released from rest, find the speed of the ring when the spring becomes horizontal. `(sin 37^@=3/5)`.

Options

A. 15 m/s
B. 10 m/s
C. 20 m/s
D. 25 m/s

Answer

C

Explanation

From work energy theorem (W=DKE)
`W_("gravity")+W_("spring")=(1)/(2)mv_(1)^(2)-(1)/(2)mv_(1)^(2)`

`mgh+((1)/(2)kx_(1)^(2)-(1)/(2)kx_(2)^(2))=(1)/(2)mv_(2)^(2)`
`(1)/(2)(10)(3)+(1)/(2)(170)(1)^(2)-(1)/(2)(170)(0)^(2)`
`=(1)/(2)((1)/(2))v^(2)-(1)/(2)((1)/(2))(0)^(2)rArrv=20m//s`