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

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`