# Mass on a spring homework

1. Nov 30, 2011

### rammer

Suppose we have a spring placed vertically on a table. We put a body with mass "m" on it. The spring compresses by "x".
Then sum of forces is zero, therefore:

kx=mg, so k=mg/x

But if we look on it as the energy problem, then: we hold the body on the string, then we release it, so loss of gravitational potential energy is "mgx" and gain of spring energy is 0.5*k*x^2.
From that:

Where I did a mistake?

2. Nov 30, 2011

### Staff: Mentor

Here you have gently lowered the mass onto the spring. Once placed in its new equilibrium position (where the spring is compressed by an amount x = mg/k), it just sits there. Note that your hand does work (negative work) on the mass as you lower it, reducing the total energy of the system.
Here you just release the mass. So all the initial gravitational PE remains to be converted to spring PE and kinetic energy, with none being removed by your hand. When it gets to the equilibrium position (at x = mg/k) it still has kinetic energy left, so it keeps going. It reaches the point x2 = 2mg/k, where the KE is momentarily zero, then comes back up. It will continue to oscillate between the two extreme positions.

3. Nov 30, 2011

### technician

This is a well known question dealing with springs !!!!
When you lower a weight onto a spring the force exerted on the spring increases from 0 up to the maximum (the weight of the object).
Therefore the work done compressing the spring = AVERAGE force x distance
= 0.5F x distance.
During the placing of a weight on a spring your hand is part of the process and ensures that the force on the spring increases from 0 to the maximum.
On the other hand (sorry !!!!!) if you release (drop) the weight then the force on the spring is constant (mg) for the compression and the extra (0.5mgh) energy appears as KE... the mass on the spring will bounce up and down

Doc Al has said the same !!!!

Last edited: Nov 30, 2011
4. Dec 1, 2011

### rammer

Thanks for both explanations :)

5. Mar 4, 2012

### superalias

I'd also like to thank Doc Al and technician for the answers. I missed this thread last month when I asked nearly the same question:

I think I get that now, thanks to the extra explanation given in this thread. If I might add a question:

I understand that when the mass is placed on the spring, displacement x = mg/k. When the mass is dropped on the spring, (non-equilibrium) displacement x = 2mg/k.

Per my question in the other thread, if we instead fired the mass horizontally at a wall-mounted spring… we can no longer talk of g, so have to look at the projectile's KE in terms of its mass and velocity. Doing so, we get: x = √(m/k) * v (momentarily, anyway, before the spring completely pushes the mass away again)

Is that correct? The confirmation would be greatly appreciated. (The difference between x proportionate to m/k when the mass is placed or dropped onto the spring, vs proportionate to √(m/k) when it's fired at the side-mounted spring, is what had me confused… but I believe I now see the misunderstanding that led to my confusion.)

6. Mar 4, 2012

### Staff: Mentor

Yes, that's correct. (Assuming no energy is lost in the collision.) That expression gives the maximum displacement from the initial position.

7. Mar 8, 2012

### superalias

Thanks for the confirmation!