# Spring potential question

josevie
Homework Statement:
An 8.00kg stone is at rest on a spring. The spring is compressed 10.0cm by the stone. (a) What is the spring constant? (b) The stone is pushed down an additional 30.0cm and released. What is the elastic potential energy of the compressed spring just before that release? (c) What is the change in the gravitational potential energy of the stoneEarth system when the stone moves from the release point to its maximum height? (d) What is that maximum height, measured from the release point?
Relevant Equations:
F=m.a =-k.x = m.g
Us = (1/2).k.x²
Ug = mgh
It is to my understanding that if the spring was compressed 10cm, it is due to the Work of the Weight Force of the stone. So:

Work done on the spring by the stone = m.g.x = 7.84 J

The work done on the spring will be stored as potential energy of the spring, so:

Us = W

Us = (1/2).k.x²
k = 2.Us/x² which gives me k = 1568 N/m, which is double the right answer of 784 N/m

I know that if I were to use Newton's second law and F(spring) = m.g, I would find the right answer. I cant, however, understand why the above method is wrong. Someone please help lol

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if the spring was compressed 10cm, it is due to the Work of the Weight Force of the stone
That would be true if the stone had been released from rest at the top of the relaxed spring. But we are told it is sitting at rest on the compressed spring, which is different.

Imagine it had been placed carefully on top of the uncompressed spring and released. With no losses, it would have compressed the spring by 20cm before coming instantaneously to rest, then bounced between the two extreme positions forever. Your calculation would apply to the work done against or by gravity over that full distance.
In practice, there are losses and the stone eventually comes to rest with the spring compressed 10cm. Half the lost GPE has gone as heat, the other half into spring PE.

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josevie
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In your energy balance, you omitted the kinetic energy which is equal to 0 when x = 20 cm, not 10.

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In your energy balance, you omitted the kinetic energy which is equal to 0 when x = 20 cm, not 10.
if this is re part a then we must be reading the question differently.

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The work done on the spring will be stored as potential energy of the spring, so:

Us = W

Us = (1/2).k.x²
k = 2.Us/x² which gives me k = 1568 N/m, which is double the right answer of 784 N/m

The spring deformation (effect) is a reaction to the applied force (cause).
Therefore, the full force of the weight is only applied onto the spring at the height of -10 cm from the point no load (full free extension).

At -5 cm, the applied force should be equal to half the weight.
At the full extension point, the applied force must be zero.

You have calculated the area of a rectangle rather than the one for a formed triangle under the reaction F versus distance curve.

The value of kappa (spring constant) is only the slope of that curve or proportionality constant of that function.

https://en.wikipedia.org/wiki/Hooke's_law

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MatinSAR and josevie
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It is to my understanding that if the spring was compressed 10cm, it is due to the Work of the Weight Force of the stone. So:

Work done on the spring by the stone = m.g.x = 7.84 J
Pointing out the error here in yet another way...

I do not know what you mean by the term "Weight Force". The contact force of the stone on the spring is distinct from the force of gravity on the stone. The problem statement makes it clear that the two will be equal only at ##x = 10 \text{ cm}##.

The work done on the stone by gravity is the product of the force of gravity on the stone and the distance moved by the stone while that force is being applied. The force of gravity on stone will be ##mg##. The distance moved is ##x = 10 \text{ cm}##. So you have a correct calculation for the work done by gravity on the stone.

The work done on the spring by the stone is the product of the force of the stone on the spring and the distance moved by the spring while that force is being applied. The force in this case will not be ##mg##. Instead, it will be given by Hooke's law applied to the spring.

To me, the "weight force" would be the force of gravity on an object. I would not use that terminology to apply to the contact force of that object on the surface below.

MatinSAR, josevie and Lnewqban
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if this is re part a then we must be reading the question differently.
I was responding to the original post

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I was responding to the original post
.. which only asks about part a. In that part there is no motion; the stone is resting statically on the spring. So what is the nonzero KE you mention?

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.. which only asks about part a. In that part there is no motion; the stone is resting statically on the spring. So what is the nonzero KE you mention?
He was doing an energy analysis as an alternative to the force balance, and I was commenting on that. At the very bottom after release at the top is the place where the KE is zero again, and elastic energy increase is equal to potential energy decrease.

SammyS
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He was doing an energy analysis as an alternative to the force balance, and I was commenting on that. At the very bottom after release at the top is the place where the KE is zero again, and elastic energy increase is equal to potential energy decrease.
Sure, but in this question it has not been "released at the top" if by top you mean the uncompressed height of the spring. It is released at the equilibrium position, so nothing moves, no KE.

josevie
That would be true if the stone had been released from rest at the top of the relaxed spring. But we are told it is sitting at rest on the compressed spring, which is different.

Imagine it had been placed carefully on top of the uncompressed spring and released. With no losses, it would have compressed the spring by 20cm before coming instantaneously to rest, then bounced between the two extreme positions forever. Your calculation would apply to the work done against or by gravity over that full distance.
In practice, there are losses and the stone eventually comes to rest with the spring compressed 10cm. Half the lost GPE has gone as heat, the other half into spring PE.
I see now my mistake of trying to use work for a system that is on rest, thanks. That, however makes me question other things:

So, if is at rest it means that -k.x = m.g, right? Does normal force has no part in it? It just came to me that I don't understand why normal force is not considered.

With it being at rest and the Net Force Fn= F(spring) + F(weight), the product of it by the distance of 10cm gives me the right answer. It would, however, be a different situation of the one mentioned above? That is, it would be situation where the spring was futher compressed 10cm over the equilibrium point? I am trying to find a interpretation on why they are different (or equal).

If it was released from the stop when the spring is not compressed, the work done on the spring would be the product of net force on the block over the distance of 20cm? To me the situation mentioned on the paragraph above seems exactly like this one and I am find it hard to understand the differences. They would however give me different values. I am finding it hard to understand if when working with energy, I would need to calculate it considering the distance from point where Kinect energy is highest or if I would need to calculate the work done by the block on the spring from the top to the lowest point.

The spring deformation (effect) is a reaction to the applied force (cause).
Therefore, the full force of the weight is only applied onto the spring at the height of -10 cm from the point no load (full free extension).

At -5 cm, the applied force should be equal to half the weight.
At the full extension point, the applied force must be zero.

You have calculated the area of a rectangle rather than the one for a formed triangle under the reaction F versus distance curve.

The value of kappa (spring constant) is only the slope of that curve or proportionality constant of that function.

https://en.wikipedia.org/wiki/Hooke's_law

View attachment 319353

That has clarified some of my question, actually. Thanks! Would you mind checking some extra questions of mine that I wrote above?

Pointing out the error here in yet another way...

I do not know what you mean by the term "Weight Force". The contact force of the stone on the spring is distinct from the force of gravity on the stone. The problem statement makes it clear that the two will be equal only at ##x = 10 \text{ cm}##.

The work done on the stone by gravity is the product of the force of gravity on the stone and the distance moved by the stone while that force is being applied. The force of gravity on stone will be ##mg##. The distance moved is ##x = 10 \text{ cm}##. So you have a correct calculation for the work done by gravity on the stone.

The work done on the spring by the stone is the product of the force of the stone on the spring and the distance moved by the spring while that force is being applied. The force in this case will not be ##mg##. Instead, it will be given by Hooke's law applied to the spring.

To me, the "weight force" would be the force of gravity on an object. I would not use that terminology to apply to the contact force of that object on the surface below.
Wait, wouldn't the force of the stone be Fnet = F(weight) + F(spring). The force would decrease as the stone goes lower due to F(spring) increasing. They are in opposite directions. Would you mind also checking some of my doubts I wrote above? Thanks!

josevie
He was doing an energy analysis as an alternative to the force balance, and I was commenting on that. At the very bottom after release at the top is the place where the KE is zero again, and elastic energy increase is equal to potential energy decrease.
Indeed, I was trying to solve using work and energy. The stone is at rest, however, so it has no Kinect energy. Is right of me to say that the system has no potential energy too? And that the only way for it to have any potential energy would be for a external force do work on the system to compress or strech the spring, thus creating an harmonic motion where the spring potential energy would be max at bottom, gravitation potential energy max at top, and Kinect energy max at the equilibrium point?

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....
That has clarified some of my question, actually. Thanks! Would you mind checking some extra questions of mine that I wrote above?
You are welcome
For example, what do you call normal force in this problem?
Why do you consider F(spring) + F(weight)?

An oscillation would follow a release from the stop when the spring is not compressed, simply because the stone would overshoot the position of equilibrium of forces (the one shown in this problem) in its way down.

MatinSAR and josevie
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Is right of me to say that the system has no potential energy too?
You have to be careful with potential energy. Where the potential energy is zero is a matter of choice. There are two forms of potential energy to consider, gravitational and elastic stored in the spring. If you choose both to be zero at the point where the mass is at rest and the spring is compressed by 10 cm, then yes, you can say that the potential energy is zero too. The answer to part (b) depends on where you choose to zero of elastic potential energy. My guess is to choose it where the spring is relaxed, i.e. 10 cm above the point where the mass is at rest and in equilibrium.

The same applies to gravitational potential energy. Fortunately, in part (c) the problem asks you to find the change in gravitational potential energy and that does not depend on the choice of zero.

MatinSAR, josevie and Lnewqban
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if is at rest it means that -k.x = m.g
Depending on how you are defining positive directions, yes. Since k, m and g are positive, that equation says x is negative, which is correct if you are defining x as the upward displacement from the uncompressed position.
Does normal force has no part in it?
Which normal force?
Taking up as positive in each case, the two forces on the mass are gravity, -mg, and the normal force the spring exerts on the mass, Nsm=-kx. Since the mass is not accelerating, these are equal and opposite: -kx=-(-mg).

That normal force and the normal force the mass exerts on the spring are action and reaction, so are equal and opposite: Nsm=-Nms.
Hence Nms=-mg.
the Net Force Fn= F(spring) + F(weight), the product of it by the distance of 10cm gives me the right answer
The net force on the mass is zero. The product with any given distance is still zero. The product of F(weight) and 10cm is the GPE lost in the stone's descent from the uncompressed spring position, but during that descent, whenever it happened, the spring force was increasing and other forces (a steadying hand) may have been present.
the work done on the spring would be the product of net force on the block over the distance of 20cm?
Since the block (released from the uncompressed position) would have accelerated at a varying rate, the net force on it was not constant. So rather than a simple product the net work done on the stone is the integral ##\int _0^{-y}(-mg-kx).dx=mgy-\frac 12ky^2## where y is the distance descended.
But that is not equal and opposite to the net work done on the spring, which is ##\frac 12ky^2##.

But the key answer to your original question is this: your energy method is invalid because in order for the stone to be resting, stationary, on the spring some other force must have been involved at some point in time.
Option 1: it was released from rest on top of the uncompressed spring. It oscillated for a while but eventually came to rest as frictional forces robbed it of KE.
Option 2: it was lowered carefully into the equilibrium position, some work being done against the hand that lowered it.
Option 3: the spring was held compressed to just the right height that its compressive force was mg, then the stone placed on it and released. The stone never moved, so no work done by gravity.

Lnewqban, jbriggs444, MatinSAR and 1 other person
josevie
Depending on how you are defining positive directions, yes. Since k, m and g are positive, that equation says x is negative, which is correct if you are defining x as the upward displacement from the uncompressed position.

Which normal force?
Taking up as positive in each case, the two forces on the mass are gravity, -mg, and the normal force the spring exerts on the mass, Nsm=-kx. Since the mass is not accelerating, these are equal and opposite: -kx=-(-mg).

That normal force and the normal force the mass exerts on the spring are action and reaction, so are equal and opposite: Nsm=-Nms.
Hence Nms=-mg.

The net force on the mass is zero. The product with any given distance is still zero. The product of F(weight) and 10cm is the GPE lost in the stone's descent from the uncompressed spring position, but during that descent, whenever it happened, the spring force was increasing and other forces (a steadying hand) may have been present.

Since the block (released from the uncompressed position) would have accelerated at a varying rate, the net force on it was not constant. So rather than a simple product the net work done on the stone is the integral ##\int _0^{-y}(-mg-kx).dx=mgy-\frac 12ky^2## where y is the distance descended.
But that is not equal and opposite to the net work done on the spring, which is ##\frac 12ky^2##.

But the key answer to your original question is this: your energy method is invalid because in order for the stone to be resting, stationary, on the spring some other force must have been involved at some point in time.
Option 1: it was released from rest on top of the uncompressed spring. It oscillated for a while but eventually came to rest as frictional forces robbed it of KE.
Option 2: it was lowered carefully into the equilibrium position, some work being done against the hand that lowered it.
Option 3: the spring was held compressed to just the right height that its compressive force was mg, then the stone placed on it and released. The stone never moved, so no work done by gravity.
That is one of the most enlightening answers I have ever read, thanks so much!!!

Lnewqban and MatinSAR
josevie
You are welcome
For example, what do you call normal force in this problem?
Why do you consider F(spring) + F(weight)?

An oscillation would follow a release from the stop when the spring is not compressed, simply because the stone would overshoot the position of equilibrium of forces (the one shown in this problem) in its way down.
@haruspex has completely annihilated any doubts I had. Thanks for taking your time to help me!

Lnewqban
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Here is may take on this from an energy perspective. The force balance equation on the mass in the upward direction is given by: $$-k(x-x_u)-mg=m\frac{dv}{dt}=mv\frac{dv}{dx}$$where ##x_u## is the unstretched length of the spring, x is the deformed length, and v is the upward velocity. If we integrate the equation with respect to x, we obtain: $$-k\left(\frac{(x-x_u)^2}{2}-\frac{(x_0-x_u)^2}{2}\right)-mg[(x-x_u)-(x_0-x_u)]=m\frac{v^2}{2}-m\frac{v_0^2}{2}$$where ##v_0## is the initial velocity of the mass when the initial length of the spring is ##x_0##.

This energy conservation equation can be rewritten as $$E=k\frac{(x-x_u)^2}{2}+mg(x-x_u)+m\frac{v^2}{2}=k\frac{(x_0-x_u)^2}{2}+mg(x_0-x_u)+m\frac{v_0^2}{2}$$ In this representation, the datum of zero potential energy is conveniently situated at the top of the unstretched spring. The total energy E is the sum of stored elastic spring energy, mass potential energy, and mass kinetic energy, and is constant.

We can use the principle of virtual work to determine the relationship between the equilibrium length of the spring and the spring constant. $$dE=(k(x_e-x_u)+mg)dx=0$$So, $$k=\frac{mg}{(x_u-x_e)}$$