Sep 7, 2013
804
172
somewhere down south
Status
Pre-Medical
So when a block on a spring is at equilibrium, the upward force due to the spring balances the downward force due to gravity, that is, kx=mg.

When is it appropriate to set the elastic potential energy of the spring (.5kx^2) equal to mg? I keep mixing up these concepts.
 
OP
orangeman25
Sep 7, 2013
804
172
somewhere down south
Status
Pre-Medical
Ok so I think I may have figured this out while I posted above? You can't set elastic potential energy to mg when spring is in equilibrium because the spring doesn't have any potential energy at that point. You also can't set elastic potential energy to mg because that's setting energy to force which doesn't make sense.
 
Dec 23, 2013
28
3
yeah elastic potential energy can be set equal to kinetic energy when the spring is pushed/pulled away from equilibrium. The kinetic energy is thus the energy that arises at it returns back to equilibrium.
you can't set it equal to mg because its in equilibrium with mg for the example that you mentioned.
I wonder if you have a spring hanging, and then its pushed up (away from equilibrium) then would 1/2kx^2 + mgh = 1/2mv^2 in order to get the speed at equilibrium
but then if its pulled down it would only be 1/2kx^2 = 1/2mv^2 in order to restore it to equilibrium.
is this right?
 
OP
orangeman25
Sep 7, 2013
804
172
somewhere down south
Status
Pre-Medical
yeah elastic potential energy can be set equal to kinetic energy when the spring is pushed/pulled away from equilibrium. The kinetic energy is thus the energy that arises at it returns back to equilibrium.
you can't set it equal to mg because its in equilibrium with mg for the example that you mentioned.
I wonder if you have a spring hanging, and then its pushed up (away from equilibrium) then would 1/2kx^2 + mgh = 1/2mv^2 in order to get the speed at equilibrium
but then if its pulled down it would only be 1/2kx^2 = 1/2mv^2 in order to restore it to equilibrium.
is this right?
Hmm I'm not sure you can use gravitational potential energy in this instance because the "h" value is just the length of the spring, and I think the spring is just the source of elastic potential energy, not gravitational. I haven't really seen gravitational potential energy being used with springs but more with pendulums
 
Jun 5, 2013
1,097
715
Status
Medical Student
The two are different terms, so you can't set them equal to each other.

KX is equal to a Newtons, 1/2Kx^2 is equal to Joules. You can set the latter equal to kinetic energy to see the terminal speed that a mass of M would have