Elastic potential energy problem

[Abu]

Homework Statement

A 1.00kg mass and 2.00kg mass are set gently on a platform mounted on an ideal spring of force constant 40.0 N/m. The 2.00 kg mass is suddenly removed. How high above its starting position does the 1.00 kg mass reach?

Related to it... An 87 g box is attached to a spring with a force constant of 82 N/m. The spring is compressed 11cm and the system is released.
A) What is the speed of the box when the spring is stretched by 7.0 cm?
B) What is the maximum speed of the box?

Homework Equations

Elastic potential energy = 1/2kx^2
Kinetic energy = 1/2mv^2
Potential energy = mgh

The Attempt at a Solution

The first problem is what confused me the most. It confused me because I don't know whether it is a case of simple harmonic motion or not. If you remove the 2 kg mass, it will move upwards, but will it come back down and oscillate or remain at the top at a new equilibrium? I will show you what I did:
mg = -kx
3(9.8) = -40x
x = -0.735
Now here is my problem, I know that you use x for the elastic potential energy 1/2kx^2, but what is the elastic potential energy equal to? Once you remove the 2kg mass, the other mass will move up. If it reaches equilibrium at the top right away, then the elastic potential energy at the bottom should transfer to just potential energy mgh, since it is not moving anymore and it is at a height? But if the mass does not stop at the top and instead comes back down, shouldn't the elastic potential energy at the bottom equal the elastic potential energy at the top?
So its either:
1/2kx^2 = mgh
OR
1/2kx^2 = 1/2kx^2 where the first x is 0.735?
I think I am confusing myself.. can someone explain this problem to me?

And for the second problem, all I need to know is if the 11 cm the amplitude? I mean the maximum displacement in the simple harmonic motion.

I know I am asking for a lot, thank you all anyways.

 

[haruspex]

the elastic potential energy at the bottom should transfer to just potential energy mgh

All of it? That would mean no elastic PE at the top. Does that seem right?

elastic potential energy at the bottom equal the elastic potential energy at the top?

So where has the gained GPE come from?

 

[Abu]

All of it? That would mean no elastic PE at the top. Does that seem right?

So where has the gained GPE come from?

So that means that elastic potential energy when it is compressed at the bottom is equal to the gravitational potential energy plus the elastic potential energy at the top?
so:
1/2kx^2 = mgh + 1/2kx^2
Does that mean this is simple harmonic motion? Or is the mass just going to reach equilibrium at the top right away?
Am i on the right track? Thank you

 

[haruspex]

So that means that elastic potential energy when it is compressed at the bottom is equal to the gravitational potential energy plus the elastic potential energy at the top?
so:
1/2kx^2 = mgh + 1/2kx^2

Am i on the right track? Thank you

Yes, but different extensions (x), of course.

 

[Abu]

Yes, but different extensions (x), of course.

Dont i have two unknowns now though? h and x on the right side of the equation?

 

[haruspex]

Dont i have two unknowns now though? h and x on the right side of the equation?

Aren't they related?

 

[Abu]

Aren't they related?

Okay. So to solve...

1/2kx^2 = mgh + 1/2kx^2
since h and k are related...
1/2kx^2 = mgh +1/2kh^2
1/2(40)(0.735)^2 = 1(9.8)h + 1/2(40)h^2
When I solve for h, i get 2 answers:
h = -1.019 m
h = 0.529 m
I am not sure if i did this right and what answer to pick? Thank you.

 

[haruspex]

1/2kx^2 = mgh +1/2kh^2

h and the maximum upward extension are related, but they are not equal.

 

[Abu]

h and the maximum upward extension are related, but they are not equal.

Oh i think I understand.
The platform, with the 3 kg total mass on it, is 0.735 m below the same platform if it had no masses on it. So we are assuming the startinf point is the platform without the masses on it right?

That means that the height above the starting point is h-x

So the new equation is:
1/2kx^2 = mgh + 1/2k(h-x)^2
1/2(40)(0.735)^2 = 1(9.8)h + 1/2(40)(h-0.735)^2
And my new height h is 0.98 meters

I saw other solutions just do
1/2kx^2 = mgh
and they got 1.1 meters. Are they wrong or am i?

 

[haruspex]

Oh i think I understand.
The platform, with the 3 kg total mass on it, is 0.735 m below the same platform if it had no masses on it. So we are assuming the startinf point is the platform without the masses on it right?

That means that the height above the starting point is h-x

So the new equation is:
1/2kx^2 = mgh + 1/2k(h-x)^2
1/2(40)(0.735)^2 = 1(9.8)h + 1/2(40)(h-0.735)^2
And my new height h is 0.98 meters

I saw other solutions just do
1/2kx^2 = mgh
and they got 1.1 meters. Are they wrong or am i?

Sorry, I blundered. I was thinking of the mass as attached to the spring, but the question says it was placed on a platform.
(I reckon my version is the more interesting question though. You answered that correctly.)

 

[Abu]

Sorry, I blundered. I was thinking of the mass as attached to the spring, but the question says it was placed on a platform.
(I reckon my version is the more interesting question though. You answered that correctly.)

Haha it's okay it happens. So which answer is the correct one then? Because I got a couple of answers now.
I have
h = -1.019 m
h = 0.529 m
and
h = 0.98 m

I'm assuming it is the first h = -1.019 m just because of the other solutions I found online? But I am not sure how come it is negative and how I would know to pick that answer instead of h = 0.529 (since I got two heights using the quadratic formula)

Sorry for dragging this on so long...

 

[haruspex]

Haha it's okay it happens. So which answer is the correct one then? Because I got a couple of answers now.
I have
h = -1.019 m
h = 0.529 m
and
h = 0.98 m

I'm assuming it is the first h = -1.019 m just because of the other solutions I found online? But I am not sure how come it is negative and how I would know to pick that answer instead of h = 0.529 (since I got two heights using the quadratic formula)

Sorry for dragging this on so long...

Neither.

This is right:

I saw other solutions just do
1/2kx^2 = mgh

 

[Abu]

Neither.

This is right:

And that's correct because when the 2 kg mass is taken off of the platform and the spring moves the 1kg mass and platform up, the platform stays at the top and reaches a new equilibrium? At this new equilibrium there is no kinetic energy because the platform is not moving. But how come there is no elastic potential energy? Is it because the spring is not trying to bring the platform back down, rather the platform just stays where it is at the top?

Thank you.

 

[haruspex]

the platform stays at the top and reaches a new equilibrium?

Once the spring is at its natural length it cannot exert an upward force on the mass, but it is not attached to the mass, so does not exert a downward force either. The mass is now a projectile. There is no force available to extend the spring, so no more spring PE generated.

One thing that ought to be checked in a thorough solution is that the value of h found is at least x. If not then there will be residual PE in the spring. This will happen if the removed mass is the lighter of the two. In this case the method in post #9 is right.

 

[Abu]

Once the spring is at its natural length it cannot exert an upward force on the mass, but it is not attached to the mass, so does not exert a downward force either. The mass is now a projectile. There is no force available to extend the spring, so no more spring PE generated.

One thing that ought to be checked in a thorough solution is that the value of h found is at least x. If not then there will be residual PE in the spring. This will happen if the removed mass is the lighter of the two. In this case the method in post #9 is right.

Thank you very much for your help and patience sir. I appreciate it very much. I believe I understand the problem as it is stated now.