# Mass on a spring

by rammer
Tags: mass, spring
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 P: 23 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: k=2mg/x, which contradicts previous equation. Where I did a mistake?
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P: 41,465
 Quote by 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
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.
 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: k=2mg/x, which contradicts previous equation. Where I did a mistake?
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.
 P: 1,506 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 !!!!
 P: 23 Mass on a spring Thanks for both explanations :)
P: 9
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:
http://www.physicsforums.com/showthread.php?t=576713

Incidentally, I received a not-as-helpful answer from the askthephysicist.com site when I inquired there:

 Your question is meaningless, the old "comparing apples and oranges" thing. The x for the carefully placed mass is the equilibrium position, the x for the dropped mass is not.
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.)
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P: 41,465
 Quote by superalias 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)
Yes, that's correct. (Assuming no energy is lost in the collision.) That expression gives the maximum displacement from the initial position.
 P: 9 Of course, an answer here just leads to more questions there… but that's for other threads down the road. Thanks again!

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