The object will drop 5 cm if released before coming to rest.
How far will the object fall?
We know from Hooke's law that the work that is done when the spring is stretched is given by the formula;
W = 1/2 kx^2
k= 100 N/m
x = 1.00 cm or 0.01 m
W = 0.5 * 100 * 0.01
W = 0.5 J
Now;
W = mgh
h = W/mg
m = 1.00 kg
g = 9.8 m/s^2
h = 0.5 J/1.00 kg * 9.8 m/s^2
h = 0.05 m or 5 cm
Learn more about Hooke's law:https://brainly.com/question/13348278
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