shm... 1

initially the spring is compressed by 3cm.. so initial PE = 10⁴*(0.03)²/2 = 4.5 J

KE of the system after the block leaves the spring = 1.v²/2

Now, from conservation of energy .. KE gained = PE lost
==> v² = 9
==> v = 3 m/s

Now .. time taken to reach the initial position where it left the spring = 2L/v = 2*0.04/3 = 0.0267 s

now.. for time taken for the block .. to stop. .. i.e. it will undergo SHM with amplitude 3cm.. at time period.. 2Ï€√1/10⁴ s = Ï€/50 s

Now.... the time taken by block .. to leave the spring initially is equal to 1/4th of an oscillation and similarly... to reach the stopping point again is 1/4th of the oscillation ...

So. total time in oscillating = T/2 = π/100 s = 0.0314s

So total time taken = 0.0581 s