What is the maximum height a weighted spring can reach?

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SUMMARY

The maximum height a weighted spring can reach is determined by the balance of gravitational potential energy (G.P.E) and elastic potential energy (E.P.E). When a mass, m, is attached to a spring with spring constant, k, and natural length, l0, the equilibrium position is given by x0 = -mg/k. To find the distance, d, from the ceiling that the ball reaches after being pulled down to an extension, x1, the potential energies at the maximum height must equal the potential energy at the release point. The equations governing this system include K.E = 0.5*m*v^2, G.P.E = m*g*(x0 + x), and E.P.E = 0.5*k*x^2.

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Homework Statement


A spring is fixed at one end to the ceiling, with an ball of mass, m, and radius, r, hanging vertically at the other end of the spring. The system is in equilibrium. The spring has constant, k, and natural length, l0.

I then pull the ball down to an extension, x1, from the equilibrium position, x0. How far will I have to pull the ball down for the ball to reach a distance, d, from the ceiling, ignoring dampening forces?

g = acceleration by gravity
x = extension at any time

Homework Equations


x0 = -mg/k
F = k(x-x0) - mg
K.E = 0 at (x0 + x1) and d
both P.E (G.P.E and E.P.E) are maximum at these points.
K.E = 0.5*m*v^2 (where v is velocity of ball)
G.P.E = m*g*(x0 + x)
E.P.E = 0.5*k*x^2
any others

The Attempt at a Solution


No idea where to start.
 
Last edited:
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Try to set the potential energy at the bottom (where the ball is released) to the potential energy at the very top.
 

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