# Calculate this Spring's Potential Energy

• EEristavi
In summary: If you follow the advice I gave you, you'll find that ##U_g=-2U_s##, you only need to plug in what you have now. Try to do it yourself, though.Another hint : ##kx^2=kxx##, ##|\vec F_g|=|\vec F_s|##.
EEristavi
Homework Statement
If we have object with m = 1 hanging on a spring and elongation is h = 0.02.
What is the potential energy of the spring after it's being stretched?
Relevant Equations
E = mgh
I know that gravitational potential energy is decreased by E = -m g h = -1 10 0.02 = -0.2. So, the spring potential energy must be E=0.2 (Joule).
However, in the answer's sheet I have E=0.1

What mistake do I make?

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The elongation is apparently 0.02 m when the mass is held by your hand and slowly allowed to reach its rest position when you release your hand. Your hand does work in this case which you have neglected. Or alternatively, if you release the mass at h = 0 , it will have kinetic energy as well as it passes h= 0.02.
Look at another approach.

I'm going to assume m = 1 kg, h = 0.02 m, and g = 10 m/s2. The force that the spring exerts on the weight is not constant. It depends on how far the spring is stretched.

The question is ill posed. The zero of potential energy can be chosen at will. Without specifying where the zero is, the question cannot be answered. To be specific about the elastic energy, in a horizontal spring-mass system the zero is normally chosen at the relaxed position of the spring; in a vertical spring-mass system the zero is normally chosen at the equilibrium position where the net force on the hanging mass is zero.

In this question we are asked to find the potential energy of the spring, not the mass, so mgh is not part of the answer.

archaic
Elastic potential energy is not the opposite of gravitational potential energy. In the case of vertical springs, however, the weight of the object and the elastic force are.
Recall that ##|\vec F_s|=kx=mg=|\vec F_g|##, that ##U_s=\frac{1}{2}kx^2## and that ##U_g=-mgx## where ##x## is the elongation of the spring. Try to find a relation between both potential energies.
Hint : ##\frac{U_g}{U_s} = ?##
NB : here we choose ##U_g=-mg\int_{h_i}^{h_f}\mathrm{dh}##.

PhanthomJay said:

I think you are making mistake here.
If hand doesn't move - displacement is 0.
A = F S = F * 0 = 0

David Lewis said:
I'm going to assume m = 1 kg, h = 0.02 m, and g = 10 m/s2. The force that the spring exerts on the weight is not constant. It depends on how far the spring is stretched.
Spring is stretched by h = 0.02 or ## \Delta x = 0.02 ##

kuruman said:
The question is ill posed. The zero of potential energy can be chosen at will. Without specifying where the zero is, the question cannot be answered. To be specific about the elastic energy, in a horizontal spring-mass system the zero is normally chosen at the relaxed position of the spring; in a vertical spring-mass system the zero is normally chosen at the equilibrium position where the net force on the hanging mass is zero.

In this question we are asked to find the potential energy of the spring, not the mass, so mgh is not part of the answer.

Maybe I translated poorly.
I'm asking what is the potential energy of the spring After it's being stretched (I will change it in the problem statement also, thanks).

So, I'm just calculating the change of potential energy - so it doesn't really matter where I take zero

Guys it's Introductory physics homework - it should be easy, don't overthink please :)
I'm just missing something really easy

EEristavi said:
Guys it's Introductory physics homework - it should be easy, don't overthink please :)
I'm just missing something really easy
If you follow the advice I gave you, you'll find that ##U_g=-2U_s##, you only need to plug in what you have now. Try to do it yourself, though.
Another hint : ##kx^2=kxx##, ##|\vec F_g|=|\vec F_s|##.
archaic said:
Elastic potential energy is not the opposite of gravitational potential energy. In the case of vertical springs, however, the weight of the object and the elastic force are.
Recall that ##|\vec F_s|=kx=mg=|\vec F_g|##, that ##U_s=\frac{1}{2}kx^2## and that ##U_g=-mgx## where ##x## is the elongation of the spring. Try to find a relation between both potential energies.
Hint : ##\frac{U_g}{U_s} = ?##
NB : here we choose ##U_g=-mg\int_{h_i}^{h_f}\mathrm{dh}##.

EEristavi
EEristavi said:
I think you are making mistake here.
If hand doesn't move - displacement is 0.
A = F S = F * 0 = 0
As noted by kuruman, the question is ill posed. I am assuming that when the mass is first placed onto the hanging spring , it is held there by your hand and slowly lowered to the point at 0.02 m where when you release your hand, the mass doesn’t move and the spring stops stretching. Your hand does work because it is displaced 0.02 m in this process. So you can’t say that spring final PE is equal to initial gravitational PE because other work is done.
You should explore the fact that if the mass is suddenly released by your hand at the very start and you immediately let go without doing work, that the spring will displace 0.04 m before it stops and then rise up again and down again until ultimately it is damped out and settles at the 0.02 m mark.

PhanthomJay said:
Your hand does work because it is displaced 0.02 m in this process. So you can’t say that spring final PE is equal to initial gravitational PE because other work is done.

This is the situation when you move your reference point or potential's zero with mass.
My reference point was not changing with respect to "Earth".
These are 2 different points of view and both are correct.

Hope you understood what I'm saying.So, This doesn't mean that my approach isn't correct.

archaic said:
##Ug=−2Us##

Ok. Now let me think: why do we have times 2 with Spring potential

archaic said:
If you follow the advice I gave you, you'll find that ##U_g=-2U_s##, you only need to plug in what you have now. Try to do it yourself, though.
Another hint : ##kx^2=kxx##, ##|\vec F_g|=|\vec F_s|##.

I now that if my answer is divided by 2 - it's correct.
I can't understand why there is 2...

archaic
Our variables : ##|\vec F_s|=kx=mg=|\vec F_g|##, ##U_s=\frac{1}{2}kx^2## and ##U_g=-mgx##.
$$\frac{U_g}{U_s}=\frac{-mgx}{\frac{1}{2}kx^2}=\frac{-mgx}{\frac{1}{2}kxx}$$
From the first equation we have ##kx=mg##.
$$\frac{U_g}{U_s}=\frac{-mgx}{\frac{1}{2}(kx)x}=\frac{-mgx}{\frac{1}{2}mgx}=-\frac{1}{\frac{1}{2}}=-2$$
Or, in other words, gravitational potential energy ##U_g=-2U_s## elastic (spring's) potential energy.

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EEristavi
Woow...
now I get it!

Very nice!
Now I have to think what's the physics behind it (this I will manage on my own).. :D

Thank you very much!

EEristavi said:
Woow...
now I get it!

Very nice!
Now I have to think what's the physics behind it (this I will manage on my own).. :D

Thank you very much!
You have to understand that the ratio of -2 is not always the case but only if the zero of potential gravitational energy is taken at the same point as the zero of elastic energy. In general,
$$\frac{U_g}{U_s}=-\frac{mg(x-x_{0g})}{\frac{1}{2}k{\left(x^2-x_{0s}^2\right)}}$$where ##x_{0g}## and ##x_{0s}## are, respectively, the points where the gravitational and spring potential energy are zero. The ratio is -2 only if ##x_{0g}=x_{0s}=0##.

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archaic and EEristavi
kuruman said:
You have to understand that the ratio of -2 is not always the case but only if the zero of potential gravitational energy is taken at the same point as the zero of elastic energy. In general,
$$\frac{U_g}{U_s}=\frac{mg(x-x_{0g})}{\frac{1}{2}k{\left(x^2-x_{0s}^2\right)}}$$where ##x_{0g}## and ##x_{0s}## are, respectively, the points where the gravitational and spring potential energy are zero. The ratio is -2 only if ##x_{0g}=x_{0s}=0##.
In my example, ##x## is the elongation, I should've written it as
kuruman said:
You have to understand that the ratio of -2 is not always the case but only if the zero of potential gravitational energy is taken at the same point as the zero of elastic energy. In general,
$$\frac{U_g}{U_s}=\frac{mg(x-x_{0g})}{\frac{1}{2}k{\left(x^2-x_{0s}^2\right)}}$$where ##x_{0g}## and ##x_{0s}## are, respectively, the points where the gravitational and spring potential energy are zero. The ratio is -2 only if ##x_{0g}=x_{0s}=0##.
Isn't the general formula for EPE ##U_s=\int_{l_0}^lk(x-l_0)dx=\frac{k(l^2-l_0^2)}{2}-kl_0(l-l_0)## if we are taking points not displacement? I think you should specify that you're taking the spring's ##l_0## point as ##x=0##.

EEristavi said:
I can't understand why there is 2...
That 2 is the same as the 2 in the formula for the area of a triangle: a = 1/2 base * height -- average width times total height. The work done against the spring as it elongates to its final extension varies from zero to the full force. The average force is equal to 1/2 the full force. The work done against the spring is equal to the average force times the distance extended.

It is also the same as the 2 in ##\int x\ dx = \frac{1}{2}x^2##

Which is, by no coincidence, the same as the 2 in the formula for potential energy of a spring, ##E=\frac{1}{2}kx^2##

Guys I understood it mathematically. However, I can't figure it out as "Change of energy concept" -
If the change of potential energy is ## mg \Delta x##, why the spring's potential energy isn't increased the same amount. As I understand we have a closed system (Or Isn't it?...).

This is the root of my mistake

Are you asking why mechanical energy is not conserved? To examine mechanical energy conservation, you need to have mass ##m## move from point A to point B. What are points A and B in this case, how does the mass move from A to B and what is the mechanical energy at each of the two points?

I have 2 Questions:

1. is mechanical energy conserved?
2. If it's not, why?Note:
As I see it's not conserved. Am I right?

EEristavi said:
I have 2 Questions:

1. is mechanical energy conserved?
2. If it's not, why?Note:
As I see it's not conserved. Am I right?

kuruman said:
What are points A and B in this case

point A - where spring isn't stretched yet.
Point B - where spring is stretched by ## \Delta x ##
kuruman said:
how does the mass move from A to B

It moves because of the gravity.
other force that is involved is force from spring F = k x
kuruman said:
what is the mechanical energy at each of the two points

if we take "starting point or 0 (zero)" to be point A:
Mechanical Energy at point A - 0;
Mechanical Energy at point B - ## mg \Delta x##
P.S. one way or another - Change of energy will always be ## mg \Delta x##

OK, that's good, but is it held by a hand and then released or is it gently lowered by the hand until it reaches the equilibrium position?

If the change of potential energy is mgΔx , why the spring's potential energy isn't increased the same amount.
I tried to explain this in my original response in post #2 and later in post #11.

Essentially, with an ideal spring, there is no way that the mass on the vertically stretched spring can be at rest in its equilibrium position B unless it is lowered there from A by the hand or some equivalent mechanical device. If it is just dropped from A without guiding it by hand, it will fly past point B, then oscillate up and down forever about B without ever coming to rest at B. Since the hand does work in moving it from A to B, mechanical energy is not conserved. If the mass is dropped from A, only then is Mechanical Energy conserved everywhere in the oscillating system, but this is not your example case. Note that mechanical energy consists of Spring and Gravity potential energies plus kinetic energy.

Rather than use energy methods, it may be best just to look at the equilibrium equation for the forces at B, then solve for the spring stiffness, k, then solve for the PE of the spring using PE = 1/2 kx^2.

EEristavi
PhanthomJay said:
I tried to explain this in my original response in post #2 and later in post #11.

Essentially, with an ideal spring, there is no way that the mass on the vertically stretched spring can be at rest in its equilibrium position B unless it is lowered there from A by the hand or some equivalent mechanical device. If it is just dropped from A without guiding it by hand, it will fly past point B, then oscillate up and down forever about B without ever coming to rest at B. Since the hand does work in moving it from A to B, mechanical energy is not conserved. If the mass is dropped from A, only then is Mechanical Energy conserved everywhere in the oscillating system, but this is not your example case. Note that mechanical energy consists of Spring and Gravity potential energies plus kinetic energy.

Rather than use energy methods, it may be best just to look at the equilibrium equation for the forces at B, then solve for the spring stiffness, k, then solve for the PE of the spring using PE = 1/2 kx^2.
I think we all agree that energy considerations are not the way to address the answer to this problem. OP seems to think that the decrease in gravitational potential energy must match the increase in elastic energy stored in the stretched spring. I was trying to get OP to see that is not the case, but in the interests of settling this, here is my reasoning.

If the mass is held in place and then released, when it passes through the equilibrium position, it will have some kinetic energy and only part of the initial gravitational potential energy is stored in the spring. By mechanical energy conservation,$$mgy=\frac{1}{2}ky^2+\frac{1}{2}mv^2.$$If the mass is held in place by a disembodied hand and brought to the equilibrium position at rest before being released by the hand, mechanical energy is not conserved. However we can apply the work-energy theorem to deal with the non-conservative force exerted by the hand, $$\Delta K=0=W_{spring}+W_{grav.}+W_{hand}=-\frac{1}{2}ky^2+mgy+W_{hand}.$$Hence
$$mgy=\frac{1}{2}ky^2-W_{hand}.$$

EEristavi and PhanthomJay
kuruman said:
OK, that's good, but is it held by a hand and then released or is it gently lowered by the hand until it reaches the equilibrium position?

It is not Stated. However, my guess is that - "It's gently lowered by the had until it reaches the equilibrium"

@kuruman @PhanthomJay

Your last comments really summarized whole idea and showed me my mistake.
I can say - I understood the concept

Thanks again very much to every person spending time and trying his best!

I guess I have no answer and we can say - Topic can be closed :)

## 1. How do you calculate the potential energy of a spring?

The potential energy of a spring can be calculated using the equation P.E. = 1/2 * k * x^2, where k is the spring constant and x is the displacement of the spring.

## 2. What is the spring constant?

The spring constant, denoted by the letter k, is a measure of how stiff the spring is. It is a constant value that depends on the material and physical properties of the spring.

## 3. How is the displacement of a spring measured?

The displacement of a spring is measured in meters (m) and is defined as the change in length of the spring from its equilibrium position when a force is applied to it.

## 4. Can you calculate the potential energy of a spring if the spring constant is unknown?

No, the spring constant is a necessary component in calculating the potential energy of a spring. It cannot be calculated solely based on the displacement of the spring.

## 5. How does the potential energy of a spring change with displacement?

The potential energy of a spring is directly proportional to the square of the displacement. This means that as the displacement increases, the potential energy also increases. However, if the displacement becomes too large, the spring may reach its elastic limit and the potential energy will decrease as the spring becomes permanently deformed.

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