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Assume I have a spring with a mass hanging from it such that the mass pulls the spring down until it just about touches the floor, but no force is applied by the mass to the floor itself.

I want to find the distance of the mass from the spring equilibrium. Since the force of the spring cancels out the force of gravity on the mass, I know that $ky = mg$, and so, the distance from the equilibrium must be $mg/k$.

However, why doesn't the principle of energy conservation work? I know that at the equilibrium, the energy is only the potential energy of gravity, which is $mgh$, and when it reaches the floor, it will only have the potential energy of the spring, and since there is no velocity, there is no kinetic energy.

And so, we get the formula:

$$mgy = \frac{ky^2}{2}$$

Which is wrong? Why?

daedsidog
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2 Answers2

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(a) By 'spring equilibrium' (your second paragraph) I think you mean unstretched (or natural) length of spring.

(b) "the mass pulls the spring down until it just about touches the floor." You need to consider this more carefully. You are also saying that the $mg =ky$ when the mass is just above floor level. This value of y is therefore the equilibrium extension. In that case if you let go of the mass when the spring was unstretched, the mass will overshoot this equilibrium position (or would if the floor didn't stop it). In other words, potential energy in the spring is not the only sort of energy that the system possesses as the mass passes through its equilibrium position ($mg =ky$) for the first time.

(c) Think about what happens subsequently. It's probably easier if the floor isn't there.

Philip Wood
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  • Thank you, it was hard for me to see that it has velocity when it reaches the floor, as it does overshoot it. – daedsidog Feb 23 '20 at 00:07
  • Good. If you've done some simple harmonic motion theory, you can work out the mass's velocity as it passes through the equilibrium position, and hence its KE. Then you can check that grav PE lost = Elastic PE gained + KE gained. Note that in practice the KE eventually gets dissipated as random energy mainly in the mass and surrounding air (and, in your case, the floor!). – Philip Wood Feb 23 '20 at 12:18
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It's better to look at the change in potential energy $\Delta U$ and the work $W$ done to the spring, as the mass descends to its equilibrium position $y=0$. Conservation then tells us: $$\Delta U+W=0\tag{1}$$

The initial position of the mass is $y$, so that:

$$\Delta U=U_{equilibrium}-U_{initial}=0-mgy=-mgy$$

We know the work $W$ done is: $$W=\frac12ky^2$$

So with $(1)$ we get:

$$-mgy+\frac12ky^2=0$$

Or:

$$mg=\frac12 ky$$ $$\boxed{y=\frac{2mg}{k}}$$ So your answer is correct.

Gert
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