Two masses (indicated as 1 and 2) of value $m$ are connected by a spring with an elastic constant $k$ and a natural length $l_0 = 0$, as shown in the figure.

enter image description here

Initially, the two masses are separated by a distance of $3d$. From that position, they are released with no initial velocity. Since $l_0=0$, the spring compresses, bringing the two masses together.

After some time, the two masses are at a distance $d$ from each other, and both have kinetic energy, as shown in the figure.

enter image description here

The principle of energy conservation states that $\Delta E = W_\mathrm{NCF}$, $\Delta K = W_\mathrm{all}$, and $\Delta U = - W_\mathrm{CF}$, where $E$ is the mechanical energy, $W_\mathrm{NCF}$ is the work done by non-conservative forces (both internal and external), $K$ is the kinetic energy, $W_\mathrm{NCF}$ is the work done by all forces (both internal and external), $U$ is the potential energy, and $W_\mathrm{CF}$ is the work done by conservative forces (again, both internal and external).

The only non-conservative force acting on this system is the normal force exerted by the floor. However, this force does no work because it is perpendicular to the direction of motion, and thus the mechanical energy must be conserved: $\Delta E = 0$.

This is not the case for the kinetic and potential energies, as one is transformed into the other. Therefore, we have $\Delta K > 0$, and there must be work done by conservative forces because, as I previously said, the normal force does no work.

The Question

The question is simple:

(Q1) Which forces do work to give the masses kinetic energy?

I have reviewed answers to other questions, but the matter is still not entirely clear to me. In particular, I found comments and ideas in the following posts:

What I've Thought So Far

(1) Considering the System $\left( 1, 2 \right)$

Things are clear in this case. The forces acting on the masses are their weights (which do no work), the normal forces exerted by the floor (which do no work), and the elastic force of the spring, as shown in the picture. All these forces are external (there are no internal forces) and both the weight and the elastic force are conservative.

In this case, the forces that do work are clearly the elastic forces that bring the masses towards the center of the spring.

(2) Considering the System $\left( 1, 2, \mathrm{spring} \right)$

If the spring is considered to be part of the system, then the elastic forces are internal and play no role in the conservation of linear and angular momentum. They do, however, play a role in the conservation of energy.

The problem is that if the spring is considered as part of the system, there are four internal elastic forces. The first one is the force that the spring applies to mass 1 $\mathbf{F}_{\mathrm{e}1}$, the second one is its reaction $-\mathbf{F}_{\mathrm{e}1}$ (the force that mass 1 applies to the spring), the third one is the force that the spring applies to mass 2 $\mathbf{F}_{\mathrm{e}2}$, and the fourth one is its reaction $-\mathbf{F}_{\mathrm{e}2}$. All of this is shown in the following picture.

enter image description here

Now, the total work done by the internal elastic forces is zero because there are equal and opposite forces on each side of the spring. Since they move in the same direction, their work cancels out.

This, however, cannot be correct because there is a change in kinetic energy, so work must be done (either by internal or external forces).

(Q2) In this picture, should the reaction forces acting on the spring be considered as part of the system? If not, this case is basically the same as the case for the other system but with the elastic forces as internal forces.

(Q3) If the elastic forces acting on the spring should not be considered, why would that be?


3 Answers 3


1.The forces which gives masses their kinetic energy are the internal restoring force of the spring which is present at the time when spring is pulled apart.These forces slowly decreases and becomes zero when spring attains its natural length. At this time masses already have some velocities. Due to this masses apply compressive force on both sides of spring. In order to resist this compressions spring also applies internal resistive force on both sides to stop the compression. These resistive forces slowly grows and decrease the velocities until they become zero. This cause the potential energy of spring to grow. And the oscillation repeats.

2.Forces are acting on spring are internal forces since they are generated by energy stored in spring.Also, Spring and masses does forms a system because external forces is zero.

  • $\begingroup$ I agree with what you state, but it still doesn't clear question 3. In the masses plus spring system, I should compute the work done by all forces (the four elastic forces that are internal to this system). When doing this, the work exerted by the four forces is zero. This, however, is not correct since there is a work done by the elastic forces. What fails, then, in this reasoning? $\endgroup$ May 31, 2023 at 16:04
  • $\begingroup$ The sign of work is indeed opposite but that doesnt stops the energy from changing its form. Spring does on mass to decrease its velocity inturn reducing its kinetic energy and mass does work on spring and increases it potential energy. Total work done is zero but individual work isn't. Newton's 3rd law verifies this. $\endgroup$
    – Alv
    May 31, 2023 at 21:12
  • $\begingroup$ But if one considers the masses + spring system, the kinetic energy has increased. Then, because $\Delta K = W_\mathrm{all}$, there must be forces that do work and therefore the total work done can not be zero. $\endgroup$ May 31, 2023 at 23:14
  • $\begingroup$ The kinetic energy of mass and spring system is constant and it is just oscillating between two extremes. Because the net external force is zero and due to nature of setup. You are looking at an individual mass but if you look at the entire system then you will find that total mechanical energy is preserved. In order to keep track of individual K.E. you need to treat masses and spring as seperate systems. $\endgroup$
    – Alv
    Jun 1, 2023 at 7:52

If you're trying to determine how energy moves, then I think you're going to confuse yourself if you try to make everything part of one system.

If the system is the masses and the spring together, then the motion of the masses and the energy in the spring are just different forms of internal energy. I would say that the kinetic energy of the system is constant because there are no external forces acting on it.

Which forces do work to give the masses kinetic energy?

If considering the masses individually, the spring. If considering the mass/spring system, nothing.

  • $\begingroup$ If no force is doing work when I consider the masses + spring system, then how can kinetic energy change at all? $\endgroup$ May 31, 2023 at 23:10
  • $\begingroup$ It cannot. In this case you should not consider the internal motion as part of the system's "kinetic energy". Just as we say a box of air sitting still has KE=0 even though the molecules in the box are moving. $\endgroup$
    – BowlOfRed
    May 31, 2023 at 23:13
  • $\begingroup$ It's slowly making sense. So, if I consider the masses + spring system, the kinetic energy (and also the potential energy) should be consider as "internal energy" and not kinetic or potential. I need to digest this a little bit... $\endgroup$ May 31, 2023 at 23:47

Now, the total work done by the internal elastic forces is zero because there are equal and opposite forces on each side of the spring. Since they move in the same direction, their work cancels out.

Neither work nor kinetic energy has spatial direction. Negative work is not work in the negative direction. Instead, it tracks whether mechanical power is being transferred to (positive) or from (negative) whatever energy repository we happen to be considering.

For a symmetrical spring mass system, the work done on both masses by the spring is positive and double the work done on either mass. (Energy is going from the spring potential to the mass's kinetic energy for bith masses.) In a completely identical statement, the work done on the spring by both the masses is negative and twice the work done by either mass.

An addendum: if you approach this from the standpoint of the work-energy theorem, properly accounting for which forces act on which object in which direction, you will find forces acting on the masses in the direction of movement (which you can integrate for positive work) and the third law reaction forces acting on the other half of each third law pair - the spring - in the direction opposite movement. Integrating will get you negative work done on the spring, which we sum with the positive work on the masses for a total of zero, conserving energy and properly showing the transfer of system energy from the configuration of the spring to the movement of the masses. I find that this approach requires extra mental effort and special care to avoid errors in thinking, but it isn't wrong.

  • $\begingroup$ Okay, then you get a total work of zero. However, the kinetic energy has increased, and therefore some forces must be doing work because $\Delta K = W_\mathrm{all}$. On the other hand, energy is not conserved because of the behaviour of elastic forces; it is conserved because the only non-conservative force (the normal force) does no work because the change in mechanical energy $E$ depends on the work of non-conservative forces. $\endgroup$ May 31, 2023 at 23:22
  • $\begingroup$ @WinterPanda That is exactly the confusion which I prefer to avoid by not taking the work-energy theorem approach as the default. Total system work is always zero for every system everywhere total work is well-defined. That's the first law of thermodynamics. That doesn't mean that the change in kinetic energy is always zero, it means that the change in kinetic energy is accounted for by an equal and opposite change in some other repository of energy. In this case, the other repository of energy is the potential energy of the spring's configuration. $\endgroup$
    – g s
    Jun 1, 2023 at 1:31
  • 1
    $\begingroup$ I think it's much more intuitively useful to just track where energy is coming from (the spring's configuration energy) and where it's going (the masses kinetic energy). $\endgroup$
    – g s
    Jun 1, 2023 at 1:33

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