Kinetic Energy of a Load Suspended by Spring

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 $$ W_k=A/2$$

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