# Spring force under gravity

Why do I have to use law of conservation of energy to solve problems regarding calculation of extension in spring length when a box attached to the lower end of the spring is released from rest (such that the spring was in it's natural length initially) and allowed to fall under gravity? I mean why can't I just simply use $$x = mg/k$$?

• What extension do you intend to calculate? – MaxWell Aug 23 '19 at 12:39
• yeah you need to specify what you are actually trying to calculate. The max amplitude? The equilibrium position? it's max velocity? etc etc – Andrew Aug 23 '19 at 12:46
• I want to calculate the maximum extension in the spring, given spring constant = k and mass of box = m. Rest information is in the question itself. Actually, by the time of me writing this comment, the answer has already been provided. – user233565 Aug 24 '19 at 7:41
• I can show you alternate ways to calculate it. If you're interested, let me know. In a comment. – MaxWell Aug 24 '19 at 19:16
• @Maxwell If you can make me understand in an alternate way, without using physics of a level any higher than that of class 12th, go on for sure. – user233565 Aug 25 '19 at 4:23

## 3 Answers

When the mass is released the total force acting on it is

$$F = mg -kx$$

in the down direction. The extension you say $$x_\text{eq}=mg/k$$ is the extension the spring has when the restoring force becomes greater than the gravity. After that point is reached, the mass continues falling until its velocity becomes $$0$$.

It is easier to solve this using conservation of energy if you say that the total energy is conserved and, at the point of the longest extension of the spring, the kinetical energy equals to zero:

$$mgx_{max} = \frac{1}{2}kx_{max}^2$$

that gives us $$x_{max}=2mg/k$$.

You can't do this because the maximum length of the spring isn't obtained when the system is at equilibrium. If you drop the block, once you do reach that equilibrium point where $$x=mg/k$$ the block is still moving. Therefore, the block from overshoots equilibrium and moves father down.

In terms of energy, at $$x=mg/k$$ the mass still has kinetic energy.

Because when the spring extends the mass losses gravitational potential energy equal to mgx that has to be accounted for.

At the maximum extension of the spring the spring potential energy equals the loss of gravitational potential energy or

$$\frac{kx^2}{2}=mgx$$

$$x=\frac{2mg}{k}$$

I should add that your equation:

$$x=\frac{mg}{k}$$

would be correct if instead of releasing the mass (so that it gains kinetic energy) you slowly lowered the mass and bring it to a stop when you no longer feel its weight, then let it go. Now you no longer need to account for an increase in kinetic energy at the expense of gravitational potential energy.

Hope this helps