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I have a physics problem in my book, where a spring is compressed and a ball is laying in the end of the spring. When the spring is released the ball will reach a certain speed. In the solution it says that all the spring potential energy will be converted into the kinetic energy of the ball.

What I don't understand is, if there was no ball, what will the energy be converted to? Heat? And if it is heat or something else, why won't we consider it when there exists a ball at the end of the string, because the string will still be moving the same way when released, however, maybe a little slower.

Thanks in advance!

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The problem in your book assumes an ideal (lossless) and massless spring. That is typical of entry level physics problems.

Without the ball you would need to consider, at a minimum, that the spring has mass that can acquire kinetic energy from the stored elastic potential energy causing it to expand beyond its initial uncompressed length until the kinetic energy is converted back into elastic potential energy, at which point the process reverses.

If the spring is additionally lossless (no heat dissipation) the spring will theoretically continue to oscillate back and forth between the fully extended and compressed lengths. If there are losses (like all real springs) the spring will eventually come to rest at its natural length.

Hope this helps.

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In the case where there is no ball (and assuming there is no friction/other dissipative forces), the spring will expand beyond its natural length due to the momentum it gains while unstretching. At some point, the expansion will reach a point where the force pulling the spring back to natural length is large enough to make the spring recoil back! It will undergo oscillations of this sort about its natural length until an external force stops it/slows it down.

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  • $\begingroup$ what is the momentum (inertia) of a massless spring? $\endgroup$
    – hyportnex
    Commented Jul 14 at 19:45
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    $\begingroup$ See Bob D's answer. In short, you cannot have a massless spring which still produces a force. In problems where we call the spring massless, we mean its mass is negligible compared to whatever it is attached to. If it is not attached to anything, the assumption breaks down. $\endgroup$
    – CompassBearer
    Commented Jul 14 at 20:07
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    $\begingroup$ exactly, and I think these considerations, especially that nothing is oscillating unless it has mass, are missing from your post $\endgroup$
    – hyportnex
    Commented Jul 14 at 20:35
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    $\begingroup$ This type of problem keeps the situation simple so students can focus on force and energy. If the spring has mass, the problem can be solved, but it is more complex. Potential and kinetic energy are distributed along the spring. Different parts of the spring are moving at different speeds. It distracts from learning the basics with $F = kx$ and $E = 1/2mv^2$. So a mass much larger than the spring is used and the spring is idealized as massless. When doing this kind of problem, you have to be careful not to change things so the idealization is broken. $\endgroup$
    – mmesser314
    Commented Jul 14 at 21:18

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