A constant force produces maximum velocity `V` on the block connected to the spring of force constant `K` as shown in the fig. When the force constant of spring becomes `4K` the maximum velocity of the block is

Options

A. V/4
B. 2 V
C. V/2
D. V

Answer

C

Explanation

By work energy theorem
`FX_(1) -(1)/(2) kx_(1)^(2)=(1)/(2)mv^(2)`....(A)
and `FX_(2)-(1)/(2) K'X_(2)^(2)=(1)/(2)mv'^(2)`....(B)
where `x_(1),x_(2)` are initial and final extensions and
`v,v'` are initial and final velocities
In both case , force applied is same, and velocity becomes maximum when `F = kx`
(after which the mass will decelerate)
`:. F = kx_(1) = (4k)x_(2) rArr x_(2) = (x_(1))/(4)`
Substituting in (B)
`(Fx_(1))/(4)-(1)/(2)(4k)((x_(1))/(4))^(2)`
`rArr (1)/(4) [Fx_(1)-(1)/(2)kx_(1)^(2)]=(1)/(2)mv^(2)`....(C)
Dividing (C)/(A), we get
`(1)/(4) = (v^(2))/(v^(2)) rArr v' = (v)/(2)`.