Question

An object of mass 1.00 kg is attached to a vertical spring with spring constant 100 N/m. The object is held at rest in a position such that the spring is stretched upward a distance 1.00 cm beyond its undisturbed length. If the object is released, how far will it drop before coming to rest

Answer 1

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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