What is the maximum height a weighted spring can reach?

AI Thread Summary
To determine the maximum height a weighted spring can reach, the equilibrium position of the spring must be established, factoring in the mass of the ball and the spring constant. The extension of the spring when the ball is pulled down is represented by x1, while the equilibrium position is x0. The potential energy at the maximum height can be equated to the potential energy at the lowest point, where kinetic energy is zero. The equations for gravitational potential energy and elastic potential energy are crucial for solving the problem. Ultimately, the solution involves balancing these energies to find the required extension for the ball to reach a specific distance from the ceiling.
arjwilliams
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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.
 
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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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