A string is suspended from a ceiling...

when a weight (500g) is attached to the lower end, it descends 10 cm and comes to rest. the weight is pulled down 20 cm, where it touches the floor. The string's force is proportional to the string's extension from its natural length. The restoring force is not exerted when the string is at less that its natural length.
When the weight is gently released from its position, what maximum heigth (from the floor) does it attain?

g=9.8 m/s^2

When the string is stretched it acts just like a spring, so we can use elastic energy (1/2)ky^2 for part of the problem.
First get the the constant "k" from the information that the mass hangs at rest when the string is stretched 10 cm;
mg = kyo
k = mg/yo = (500)(980)/10 = 49000 dynes/cm

When the string is stretched 30 cm to the floor its potential energy , relative to y=0 at no stretch position, is;
PE = (1/2)ky^2 - mgy

When the mass is released, all of this PE is equal to KE when the mass reaches the no stretch position of y=0;
(1/2)mv^2 = (1/2)ky^2 - mgy

The mass will continue moving up to some maximum height "h" ,above the y=0 position, where all of its KE becomes gravitational PE
mgh = (1/2)mv^2

So
mgh = (1/2)ky^2 - mgy
h = ky^2/2mg - y

This is the height above the no stretch position. The height above the floor is then;
H = h + y
H = ky^2/2mg
H = (49000)(900)/2(500)(980)
H = 45 cm