A Block of mass m = 0.1kg  is connected to a spring of unknown spring constant k.  It is compressed to a distance x from its equilibrium position and released from rest.   After approaching the distancex2   from equilibrium position , it hits another block and comes to rest momentarily, while the other block moves with a velocity 3m/s.   The total initial Energy of the spring is,

The block comes to rest after collision ⇒block velocity before collision was 3m/s, before collision, KE of the spring block system is,  K.E   =    12 mv2  substitute given values ⇒KE =   12 0.1×32 ⇒KE   =    0.45 Joule---(1)     P.E.   = 12kx2 at displacement=x2 , PE= 12k(x2)2---(2) Total energy =12kA2 at displacement x, TE=12kx2---(3) substitute equation(3) in equation (2) PE=14TE         then remaining energy will be KE KE=34TE substitute equation(1)  0.45 =34TE TE=0.45 X43 Total energy =0.6 Joule  ∴   T.E   =    43×0.45   =    0.6 J