Conservation of Energy of a Spring with Weight Added

[The Head]

Homework Statement

Please see attachment-- it's problem #35

Relevant Equations

kx=mg, P_g=mgh, P_e=1/2kx^2, conservation of energy equations

I'm all messed up on this problem. I see you can get the solution (74cm, as listed in the back of the book) by simply setting mgh=1/2kx^2, saying that h=x, and then adding 15 cm to that since that's the original length of the spring. This is the solved solution I was given. But now I think it's more complicated than that and have gotten really tangled and confused.

There's a new equilibrium position of the spring when you add the 2.5 kg mass, and that's easy to find, since kx=mg. So it sort of makes sense that the lowest point would be twice delta_x since all that gravitational potential energy converted to kinetic energy by the time equilibrium was reached, and then stretches to elastic potential energy. But isn't there also now elastic potential energy in the beginning? That is, if the new equilibrium spot is mg/k below the original, then at the original spot the spring is actually slightly compressed.

So for equations, I'm thinking:
P_g1 + P_e1 = P_e2

If I make my "zero PE point" equal to the new equilibrium point, then I have both gravitational and elastic potential energy at the initial position and then similarly both of these forms at the bottom (but gravitational is negative)

mgx_1+ 1/2k(x_1)^2=1/2k(x_2)^2 -mgx_2, where x_1= mg/k

2.5(9.8)(0.295) +1/2 * 83(0.295)^2 = 1/2 * (83)(x_2)^2- 2.5(9.8)(x_2)

Solving for x_2, I get x_2 = .88 and -.295 and choose the negative root

My question is, why does it work to simply say 1/2kx^2=mgx, when the x value in the P_e (amount stretched from equilibrium at it's lowest point is actually not the same as the x value in P_g (the change in height from the high to low point).

 

[haruspex]

elastic potential energy in the beginning? That is, if the new equilibrium spot is mg/k below the original, then at the original spot the spring is actually slightly compressed.

Not sure what you are saying. By "beginning" and "original spot" you mean before the mass is released, yes? By definition, there is no elastic PE at that point. Why do you think there is?

 

[The Head]

Not sure what you are saying. By "beginning" and "original spot" you mean before the mass is released, yes? By definition, there is no elastic PE at that point. Why do you think there is?

The problem has a spring at equilibrium, and then a mass is added to it, so I assumed that means it‘s no longer at equilibrium (it would have a new equilibrium point determined by kx=mg, right?). Thus, if it‘s not at equilibrium, at the point of release, it would have elastic potential energy because delta_x doesn‘t equal zero. Is that incorrect?

I also realized in my set-up, I should choose the positive root because I assumed mgx_2 was negative, which apparently gives me the wrong solution. Aghhh!

 

[haruspex]

it‘s not at equilibrium, at the point of release, it would have elastic potential energy

No. Before the mass is attached, no elastic PE.
When mass is attached but not yet released, no change in the string, so still no elastic PE.
After the mass is released the string gains elastic PE, the mass gains KE and loses GPE.
At the equilibrium point, KE is maximum.
Beyond that, elastic PE continues to increase, GPE continues to be lost, and KE reduces.

 

[The Head]

No. Before the mass is attached, no elastic PE.
When mass is attached but not yet released, no change in the string, so still no elastic PE.
After the mass is released the string gains elastic PE, the mass gains KE and loses GPE.
At the equilibrium point, KE is maximum.
Beyond that, elastic PE continues to increase, GPE continues to be lost, and KE reduces.

Thanks for your response and for outlining the energy forms. It makes sense that when the mass is released the spring stretches and gains elastic PE conceptually. But I thought the definition of the equilibrium point was when the elastic PE=0. I guess that’s only in horizontal situations, and we’re saying at equilibrium there is in fact elastic PE?

And so delta_x would not be zero at the equilibrium point? Then I guess the equation is -mgx=1/2 kx^2 + 1/2 mv^2 where x is the change in position from where it is released?

 

[haruspex]

I thought the definition of the equilibrium point was when the elastic PE=0.

No, it's the position in which forces balance, so there's no acceleration.

 

[The Head]

No, it's the position in which forces balance, so there's no acceleration.

Thanks. And in this case the ‘x’ in 1/2kx^2 isn’t the distance from equilibrium? And the other statements were OK?

 

[haruspex]

the ‘x’ in 1/2kx^2 isn’t the distance from equilibrium?

Indeed. If a string of relaxed length L is stretched by x the EPE is ½kx². If it is then stretched by an additional Δx the increase in EPE is ½k(x+Δx)²-½kx²=½k(2x+Δx²).