Why is the elastic potential energy when extension is (a+l/20) included?

[Janiceleong26]

1. Homework Statement

I'm working on the second part of this question and I want to find the initial speed before P strikes the barrier.
So I used principle of conservation of energy,
K.E at Equilibrium position + elastic p.e. at Equilibrium positon = K.E just before it strikes the barrier + G.P.E at the barrier

But according to the solutions,

Why Is the e.p.e when extension is (a+l/20) included? I thought the string is cut, so there shouldn't be any tension anymore

Homework Equations

The Attempt at a Solution

 

[PeroK]

If the mass is hanging on the string, then there will be tension in the equilibrium position. The tension in the string must balance the gravitational force on the mass.

 

[haruspex]

The KE when descending through the equilibrium position will be the same as when ascending through it.
It starts with no KE, but elastic PE corresponding to an extension of a+l/20, a being the equilibrium extension. In rising to the equilibrium position, it gains mgl/20 in GPE, but its extension decreases to a. Thus the gain in KE is EPE(a+l/20)-EPE(a)-GPE(l/20).

 

[Janiceleong26]

If the mass is hanging on the string, then there will be tension in the equilibrium position. The tension in the string must balance the gravitational force on the mass.

The KE when descending through the equilibrium position will be the same as when ascending through it.
It starts with no KE, but elastic PE corresponding to an extension of a+l/20, a being the equilibrium extension. In rising to the equilibrium position, it gains mgl/20 in GPE, but its extension decreases to a. Thus the gain in KE is EPE(a+l/20)-EPE(a)-GPE(l/20).

Ok thank you.