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A heavy load is suspended on a light spring with upper half rigidity $$2k$$. The spring is slowly pulled down at the midpoint (a certain work $$A$$ is done thereby) and then released. Determine the maximum kinetic energy $$W_k$$ of the load in subsequent motion.

Solution:

If the middle of the spring is stretched out by a distance $$x$$ while doing work $$A$$, the entire spring is stretched out by $$x$$.
Hence, the potential energy of the spring which is equal to the kinetic energy in the subsequent vibrational motion is $$W_k=kx^2/2$$
When the spring is pulled downwards at the midpoint, only its upper half (whose rigidity is $$2k$$) is stretched, and the work equal to the potential energy of the extension of the upper part of the spring is $$A=2kx^2/2=kx^2$$. Hence, we may conclude that the maximum kinetic energy of the load in the subsequent motion is $$W_k=A/2$$