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

Answer

`V//2`

Explanation

By work energy theorem,
`w_("ext") + w_("spring") = k_(f) - k_(i)`
Let `x_(1), x_(2)` be the initial and final extensions and
`v, v'` initial and final velocities
initially,
`Fx_(1) - (1)/(2)kx_(1^(2)) = (1)/(2) mv^(2) …..(1)`
and finially `Fx_(2) - (1)/(2) k^(')x_(2^(2)) = (1)/(2) mv^('2) .....(2)`
In both cases : force applied is same, and velocity become maximum when `HF = kx`. (at equilibrium)
(after which the mass will deaccelerate)
`:. F = kx_(1) = (4k)x_(2) rArr x_(2) = (x_(1))/(4) , (k' = 4k)`
Substituting in `(2)`
`fx_(2) - (1)/(2)k' x_(2^(2)) = (1)/(2)mv^('2)`
`(Fx_(1))/(4) - (1)/(2)(4K)((x_(1))/(4))^(2) = (1)/(2)mv^(2) rArr (1)/(4)[Fx_(1) = (1)/(2)kx_(1^(2))] = (1)/(2) mv^(2) ...(3)`
Dividing `(3)//(1) :` we get :
`(1)/(4) = (v^(2))/(v^(2)) rArr v' = (v)/(2)`