Isn't the work done on a spring-mass system zero, so that there is be no change in potential energy?

Question

So far what I had understood about potential energy is that it is defined for a system of particles (at least two particles) with forces acted by the particles on each other of same magnitude and opposite in direction as, $$-W_\mathrm{conservative} = ΔU.$$

For instance in case of gravitational potential energy, we only consider work done on small body as there is negligible work done on the earth. But in cases such as the following example, the concept is not completely clear. Consider a massless spring, the left side of which is attached to the wall, and there is a block attached to the right side of the spring. If we consider the block $\cup$ spring as our system, then the the net work done by internal forces would be zero. So what is increasing the potential energy of the system as we stretch the spring?

As per one of the answers to this question, it seems we need to consider the earth to be a part of the system and the spring just stores potential energy similar to the gravitational field while not really being the part of the system.

Is this reasoning correct? If not, please help me clarify the concept.

Answer 1

If we consider block + spring as our system then the the net work done by internal forces would be zero , So what is increasing the potential energy of the system then ?

See this is the whole point behind the harmonic oscillator , you don't need any 'external' to drive the object to keep it's oscillating behaviour. You give the harmonic oscillator a little push and, due to the nature of the set up, the force law is such that that the body keeps moving without ever stopping (in the ideal case).

As per 'what' is increasing the potential energy, you can say that the internal forces of the system are responsible for it. The internal forces do no work, but it needn't be that they can't convert PE and KE back and forth.

I think it may help to learn the distinction between gravitational potential and gravitational potential energy see here

Answer 2

You need to provide the potential energy by doing work on the spring-mass system to stretch or compress the spring.

If you stretch or compress the system by a distance x, the restoring force on the block by the spring will be $-k x$. Hence in order to stretch/compress the spring, the force that you apply has to be equal and opposite to this force and that will be $+k x$. If you pull the spring from the mean position = 0 to a position x, then the work done by you is $\dfrac{1}{2}$ k $x^2$. This work done gets stored as the potential energy and when you release it, the mass will start moving. At that point there is no longer any external force and the potential energy starts getting converted to kinetic energy. This conversion goes back and forth. The source of the energy however, is the external force.

The energy of the spring mass system that is equal to $\dfrac{1}{2} k A^2$, where A is the amplitude of the oscillation is thus provided from an external source. This can be done in many different ways. However you have to provide the system with kinetic or potential energy initially so that it has some energy. This can then convert between kinetic and potential energy.

The same situation (w.r.t. work done and potential energy) also occurs when you place a body at a height in a gravitational field. That work-energy phenomenon also follows the same logic. Work is done by an external force is lifting the object to the height and this is then stored as the gravitational potential energy ($mg \cdot \text{height}$).

Answer 3

For the elastic potential, the wall plays the role of a big celestial body in the case of gravitational potential.

In both cases, one of the objects are supposed to be big enough to be taken as fixed. And the other moves according to the field.

The earth alternates periods of greater distance from the sun, (higher potential) and smaller kinetic energy, and others of smaller distances and greater kinetic energy.

The spring also oscillates with a trade-off between potential and kinetic energy. The elastic field $\frac{1}{2}k\Delta x^2$ of the spring can be compared to the gravitational field $-\frac{GM}{r}$.

Answer 4

I was able to clear my misconception , So I am just writing so anybody else could get helped ;

So first of all , I got to know that only "internal" conservative forces changes potential energy as: $$-W_{internal} = ΔU \tag{1}$$ Note:- I am considering non-conservative internal forces to be zero.

Note when we write Work energy theorem , the $W_{net}$ includes both $W_{internal}$ and $W_{external}$. $$W_{internal} + W_{external} = ΔKE$$

By using $(1)$ $$W_{external} = ΔU + ΔKE \tag{2}$$

If $W_{external}$ is zero then we say total energy of system is conserved.

Now let's consider spring as our system in which let $W_{internal}$ mean work done by internal particles of spring on each other due to elastic forces and the block applies the force $kx$ on the spring which is $W_{external}$. As spring is massless this implies $ΔKE$ = 0 ,So we could write work energy theorem for the spring as $$W_{internal} + W_{external} = 0$$ Using (1) $$W_{external} = ΔU_{spring}$$

By calculation , $W_{external}$ turn outs to be $kx^2/2$. $$ΔU_{spring} = kx^2/2$$

Now if I were to consider Block ∪ Spring as my system then: $$W_{block on the spring} + W_{spring on the block} = 0$$

Considering no external force on the block ; $$W_{internal} = ΔKE$$ By using (1) $$ ΔKE + ΔU = 0 $$

So my statement "If we consider the block ∪ spring as our system, then the the net work done by internal forces would be zero" was wrong as there are internal forces of spring which are responsible for potential energy !!

I would also like to clear out that in a field the source remains fixed , So potential energy of the "source + particle" system is wrongly attributed as potential energy of particle as source has $ΔKE = 0$.

Reference : - HC Verma : Concept of physics - Volume 1