At `t = 0`, a constant force is applied on `3 kg` block. Find out maximum elongation in spring in cm.

Answer

`8 cm`

Explanation

Given system can be reduce by using reduced mass concept `mu = (2 xx 3)/(2 + 3) = (6)/(5) kg`
and `F_("reduced") =` force on any block w.r.t. centre of mass `= (m_(1)F_(2) + m_(2)F_(1))/(m_(1) + m_(2)) = (2 xx 10 + 3 xx 0)/(2 + 3) = 4N`
`(1)/(2) kx^(2) = 4x rArr x = (8)/(k) = (8)/(100) = 8 xx 10^(-2)m = 8 cm`