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?

Question

PE + KE = PE + KE

Answer 1

initial PE=final PE

let the starting point be x=0
initial spring PE=1/2 k 1^2
initial gpe=0

Initial PE=final PE(spring and gpe)+finalKE
1/2 k 1^2=1/2 k (x-1)^2-mgx

100=100(x-1)^2-2*1*9.8x
100=100(x^2-2x+1)-19.8x

x^2-1.802x -1=0 check that.

quadratic equation
x=(1.802 +- sqrt (1.802^2+4)/2

x= you do it. It is too easy to err of these typing.