A block of mass `m = 2 kg` is connected to a spring of force constant `k = 50 n//m`. Initially the block is at rest and the spring has natural length . A constant force `f = 60 N`, is applied horizontally towards right, the maximum speed of the block (in `m//s` will be (neglect frication).

Answer

6

Explanation

For maximum speed.
`(dv)/(dt) = 0 rArr a = 0`
It at maximum speed, elongation is `x`, then
`F = kx rArr x = (F)/(k)`
Now, by `WET`.
`-(1)/(2)kx^(2) + Fx = (1)/(2) mv^(2)`
`V = (F)/(sqrt(mk)) = 6 m//s`