Work-Energy Theorem on a spring with mass

Question

Suppose you a have a spring with a certain mass (in the sense that the spring itself has mass, not in the sense that the spring has a mass attached to the movable end) that is compressed some distance and then let go of. One end of it is fixed to a wall.

Suppose for this question that the spring is horizontal, so gravity doesn't do any work, and that the spring isn't subject to friction or air resistance as it moves.

As far as I can tell, the only external force acting on the spring in the direction that it moves is the force from the wall, keeping the fixed end fixed. However, since that force is applied to a point that doesn't move, it doesn't do any work on the spring.

So, by the work-energy theorem, the change in the spring's kinetic energy should be zero, but as it returns to its uncompressed state, it should have some kinetic energy. Where have I gone wrong in my calculation?

Answer 1

The only external force acting on the spring-mass system as a whole is the force that keeps the fixed end stationary. As you say, the point of application of this force does not move, so this force does no work on the system. Thus the total energy of the spring-mass system is constant.

However, if we consider the spring and the mass separately, then we need to consider the tension in the spring (when considering the whole spring-mass system, the tension in the spring is an internal force, which is why we ignored it up until now). As the mass moves to and fro, the tension in the spring does work on the mass and the mass in turn exerts a force on the spring. Newton's Third Law tells us that force exerted by the spring on the mass and the force exerted by the mass on the spring are equal and opposite.

So the work done by the spring on the mass (call this $W_1$) and the work done by the mass on the spring (call this $W_2$) sum to zero (which is why the total energy of the spring-mass system is constant). However, $W_1$ and $W_2$ are not individually zero - they just have to sum to zero. When the tension in the spring and the velocity of the mass are in the same direction then $W_1 > 0$ and $W_2 < 0$, the kinetic energy of the mass is increasing, and the potential energy stored in the spring is decreasing. When the tension in the spring and the velocity of the mass are in opposite directions then $W_1 < 0$ and $W_2 > 0$, the kinetic energy of the mass is decreasing, and the potential energy stored in the spring is increasing.

Answer 2

I assume you are considering a spring with some mass and having a constant mass distribution. The spring is compressed some distance $x$. Now, consider the end of the spring and take a very small length $\delta l$ at this end. Since the spring has a uniform mass distribution, this $\delta l$ has some small amount of mass $\delta m$. Now the spring would behave as if $\delta m$ amount of mass is attached to its end and the compression would store potential energy. In the same way, we can take the length and mass of the entire spring in small bits and with integration, get the net potential energy stored. Without dissipation, we can apply now the Work-Energy theorem.

As far as the wall is concerned, the force exerted by the wall is just the normal reaction of the force of compression that you are applying on the spring. It is always equal and opposite to the force of compression. Also, the normal force from the wall is opposite to compression (also extension), the work done by it is negative and is stored as the potential energy. I hope my answer would help you.

Answer 3

Sorry for the confusion, everyone, I think I figured it out.

I was trying to apply the work-energy theorem as presented in my textbook, that the total work done on a particle by external forces is equal to the change in its kinetic energy.

The spring, of course, isn't a particle, and cannot be represented as one since different parts of it are moving at different speeds. The book explicitly mentions that the theorem as presented cannot be applied to composite systems like the spring.

The potential energy angle works and gives the correct answer, but that's in the next chapter and I was trying to solve a question using only the tools of the chapter itself.

Regardless, thank you for helping me out.