Asked by Joe
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?
PE + KE = PE + KE
PE + KE = PE + KE
Answers
Answered by
bobpursley
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.
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.
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