If a spring is stretched by a weight of mass m, (so the extension is $\Delta x$) then $ k = \frac W{\Delta x} = \frac {mg}{\Delta x}$. So $ k\Delta x= mg $.

When the Spring is stretched by distance $\Delta x$ (by the weight) then it looses gravitational potential enegry. ($ \Delta GPE= mg\Delta x$)

But, when we calculate the change in elastic potential energy, We get
$ U = \frac 12 k(\Delta x)^2 $. Since $ k\Delta x = mg$, $U=\frac 12 mg\Delta x$

At, equilibrium we have no kinetic energy $E_k = 0$

Doesn't that violate Energy Conservation?

Where does the rest of lost GPE goes?

Here's a illustration up.jpg


3 Answers 3


If you attach the weight and let it drop, it will fall and gain kinetic energy, overshoot the equilibrium point, slow down, change direction, and then cycle over and over again. In order for it to ever reach equilibrium, the energy must dissipate as heat via air resistance, friction in the spring, etc.


This is a subtle matter. The block needs to move quasi-statically to keep its kinetic energy zero, and for that reason we would need an external force F. This force will also do negative work on the block, reducing its change in GPE by half, which is stored in the spring. If the process was not quasistatic, the block would have a velocity after travelling some distance, and your question is answered, since you need to take the kinetic energy then as well. Note that F is not a constant force, because it has to counter the effect of gravity as well as the changing spring force.


Look at the force $F$ vs extension $x$ graph for a spring.
It is a straight line graph through the origin of gradient $k$ the spring constant and $F=kx$.
The work done by the external foce $F$ to extend the spring from being unextended, $x=0$ until it has an extension $x_o$ is $\displaystyle \int_0^{x_o} F dx = \int_0^{x_o} kx \;dx = \frac 1 2 k x^2$
Put another way.
The average force during the extension is $\dfrac {kx_o} {2}$ and so the work done by the external force is $\dfrac {kx_o} {2} \; x_o = \dfrac {kx_o^2} {2}$

Now when you add a mass $m$ to the end of the spring that mass has a constant weight $mg$ and so potentially can exert a constant force on the spring.
You can replicate the analysis done above with a force on the spring which changes with the extension of the spring by applying an upward force $F_{up}$ on the mass such that the net force exerted on the spring $F = mg - F_{up}$ and you will then get that the energy stored in the spring is $\dfrac 1 2 k x^2$ as the work done on the spring is $\displaystyle \int_0^{x_o} (mg - F_{up}) \; dx = mg\;x_o + \left[ - \int_0^{x_o} F_{up} \; dx \right ]$

The first term being the work done by the gravitational force and the second term being the work done on the force $F_{up}$

If the force $F_{up}$ is not present then the $mg$ again does work $mgx_o$ but now the mass $m$ is accelerating since $mg > kx$ and it carries on accelerating until $mg=kx_o$ when the net force on the mass is zero.
However although this is the static equilibrium condition in terms of forces the mass is moving having gained kinetic energy $\dfrac 1 2 k x_o^2$ during its descent will continue onwards until it finally stops when the extension $x = 2 x_o$.
In terms of energy, the spring has stored in it a potential energy of $\dfrac 1 2 k (2x_o)^2 = 2 kx_o^2$ in it and the work done by the gravitational force is $mg\;2x_o = 2 kx_o^2$.

So no energy has been lost.

If the spring mass-system were left alone and there were no dissipative forces acting then the mass would oscillate about the static equilibrium position $x=x_o$ for ever.

In practice with frictional forces present the mass would undergo damped harmonic motion and eventually finish up stationary at the static equilibrium position with energy $\dfrac 12 k x_o^2$ dissipated as heat.

  • $\begingroup$ What if we use the initial and final conditions only.M is gently loaded so that kinetic energy does not come into the picture.Once at equilibrium the spring has gained the potential energy lost by m ie mgx = 1/2kx^2. Also at equilibrium mg= kx.Both are standard equation used for problem solving.But if we replace mg= kx in the energy balance equation LHS not = RHS.Why? $\endgroup$
    – Chappy
    Dec 14, 2017 at 18:27
  • $\begingroup$ @Chappy To keep the mass from gaining kinetic energy there must be an upward external force on the mass in addition to the upward force due to the spring. Note that the net force on the mass must be zero. When the mass drops down a distance $x$ the spring gains an amount of energy equal to $\frac 12 k x^2$ and the work done on the object creating the external force is also $\frac 12 k x^2$ so you have $mgx = \frac 12 k x^2 + \frac 12 k x^2 = k x^2 \Rightarrow mg =kx$. $\endgroup$
    – Farcher
    Dec 14, 2017 at 20:28
  • $\begingroup$ while solvong problems say damping while bungee jumping etc we take the loss of porential energy equal to 1/2 kx^2 and assume that while the man is in free fall his difference of potential energy converts to kinetic energy but once the string starts stretching both these convert to elastic potential energy.Since at the beeginning total energy was purely potential energy we equate mgx = 1/2 kx^2.In case half of the change of potential energy is dissipated isn't this approach incorrect?And in case it doesnt dissipate and the equation ir correct then mg= kx gives LHS not = RHS. $\endgroup$
    – Chappy
    Dec 15, 2017 at 1:44
  • $\begingroup$ youtu.be/XMX8KDHD5f4 $\endgroup$
    – Chappy
    Dec 15, 2017 at 3:01
  • $\begingroup$ @Chappy when you write $mgx = \frac 12 k x^2$ ie equating the loss in gravitational pe to the gain in spring pe you must realise that the net force on the mass is not zero. When you write $mg=kx$ then the net force on the mass is zero. This shows that the two $x$s are not the same. A net force of zero does not necessarily mean that the mass is stationary ie has no kinetic energy. I am not sure what the video is supposed to show me. $\endgroup$
    – Farcher
    Dec 15, 2017 at 7:36

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