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)=0impliesa=0`
if at maximum speed elongation is x, then
`f=kximpliesx=(F)/(k)` now by WET,
`-(1)/(2)kx^(2)+Fx=(1)/(2)mv^(2)`
`V=(F)/(sqrt(mk))=6(m)/(s)`