How to assign signs to energy?

[hdp]

Homework Statement

A mass M is supported by a light spring of constant k. Calculate the amount of mechanical energy if the mass is at rest.
(The mass is now at height h)

Relevant Equations

Spring potential = 0.5kx^2
Gravitational potential = mgh
As it is at rest :
Kx=mg

The gravitational potential leads to velocity in downward direction, but spring potential does in upward direction. So should these energies have different signs (plus and minus or vice-versa)?

 

[kuruman]

Energy is not a vector and does not have a direction.

Your equation kx = mgh is incorrect because it sets a force equal to an energy.

 

[BvU]

Homework Statement: A mass M is supported by a light spring of constant k. Calculate the amount of mechanical energy if the mass is at rest.
(The mass is now at height h)

Is this the exact wording of your exercise ?

Relevant Equations: Spring potential = ${1\over 2} kx^2$

Gravitational potential = $mgh$
As it is at rest :
Kx=mg

The gravitational potential leads to velocity in downward direction, but spring potential does in upward direction. So should these energies have different signs (plus and minus or vice-versa)?

So interesting: the information you have been given in this version of the exercise is not very consistent; not with itself and not with what you then do with it.

The exercise: 'The amount of mechanical energy' of what ? Compared to what ?
'The mass is now at height h' with respect to what reference point where $h = 0$ ?

Let me paint a picture in three panels:

In A we have the spring 'at rest'. For later reference I added a reference level $x=0$.
And additionally I define a positive direction 'up' for x, coincident with the positive direction for $h$.

In C we have the spring + mass at rest on it.
'at rest' meaning the mass is not moving wrt the floor.

In between I drew B where the mass is on top of the spring, not yet compressing the spring but held in place by some upward force $F = Mg$ (The uppercase M you use)

How to get from B to C (where the upward force $F = mg$ is now entirely due to the spring -- so with your formula you have $k\Delta x = Mg$ ) ?

You have $\Delta E_G = Mg\Delta x$ on the one hand and $\Delta E_{\text spring} = {1\over 2} k\Delta x^2$

How can these two be matched ? After all $k\Delta x = Mg\ \Rightarrow \ k\Delta x^2 = Mg \Delta x$ but then what about the $\Delta E$ ?

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[Steve4Physics]

Hello @hdp. Welcome to PF.

In addition to @kuruman's, @Delta's and @BvU's comments....

but spring potential does in upward direction.

No. It has no direction. E.g. consider an 'open-coil' vertical spring - one that can be stretched or compressed. It is fixed at the top, with a mass attached at the bottom.
When the spring is in a stretched condition, it will tend to pull the mass up.
When the spring is in a compressed condition, it will tend to push the mass down.
But the spring has positive elastic potential energy in both cases.

So should these energies have different signs (plus and minus or vice-versa)?

Kinetic energy (KE) and elastic potential energy (EPE) are always positive.

For gravitational potential energy (GPE) you choose where the reference-level is (e.g. ground level). Above the reference-level, masses have positive GPE. Below the reference-level, masses have negative GPE.

In your question, it sounds like all values of height ($h$) are above ground level - so GPE is always positive in your question. (Assuming there are no holes in the ground for things to fall into!)

But a change in energy can be positive (meaning an increase) or negative (meaning a decrease). For example, when you drop a ball, the change in KE is positive but the change in GPE is negative.

 

[kuruman]

Homework Statement: A mass M is supported by a light spring of constant k. Calculate the amount of mechanical energy if the mass is at rest.
(The mass is now at height h)

In my opinion the formulation of this problem does not provide enough information. All forms of energy have an arbitrary zero. Here, we can reasonably agree that the kinetic energy is meant to be zero because the mass is said to be "at rest" although no with respect to what.

However, we are not told where the zeroes of gravitational and elastic potential energy are. Specifically, for a vertical spring, as is the case here, it is expedient and quite common to choose the zero of both kinds of potential energy at the equilibrium point which makes it possible to write the potential energy when the mass is displaced by $x$ from equilibrium as $U=\frac{1}{2}kx^2$ and ignore the gravitational part.

Considering all that and in the absence of given zeroes of energies, I would argue that the mechanical energy of the system is zero.

 

[kuruman]

The elastic potential energy is always positive? According to the well known definition that holds for all kinds of potential energy (post #9) and by taking $U_A=0$ we can see that the sign of the $U_B$ is the same as the sign of the work $W_{AB}$? Why is that work always positive?

The potential energy function $U(x)$ is defined as the negative of the work done by the conservative force from some reference point to an arbitrary point $x$. For a spring, the potential energy function is $$U(x)-\cancel{U(x_{\text{ref}})}=-\int_{x_{\text{ref}}}^x F(x)~dx=+k\int_{x_{\text{ref}}}^x x~dx=\frac{1}{2}k\left(x^2-x_{\text{ref}}^2\right).$$It is always positive if both of these requirements are satisfied
(a) The variable $x$ is zero when the spring is relaxed, i.e. $x$ is the coordinate of the mass.
(b) The reference point $x_{\text{ref}}$ is set at $x=0.$

 

[haruspex]

In my opinion the formulation of this problem does not provide enough information.

this boils down to a matter of conventions

The only reasonable interpretation of the question is that it is asking for the amount of readily available mechanical energy remaining. That does not depend on any convention and there is enough info.

 

[Delta2]

The only reasonable interpretation of the question is that it is asking for the amount of readily available mechanical energy remaining. That does not depend on any convention and there is enough info.

I haven't heard before this combined concepts "readily available" what exactly do you mean by that

 

[haruspex]

I haven't heard before this combined concepts "readily available" what exactly do you mean by that

What would you do to extract mechanical energy from the arrangement described?

 

[haruspex]

Where am i wrong here??

First, define what your U represents. For this question, we want the total mechanical energy, so the system should include mass, spring and Earth.
While the spring is being compressed or stretched, whichever, it does negative work on the rest of the system, but that equals the EPE it gains. No net change to U there.

But we don't care what the energy transformations were to reach the stated arrangement, only what we can then extract.

 

[Steve4Physics]

Hope this pulls together a few issues...

There is an explicit reference to the height, $h$, of the mass. It seems reasonable to interpret $h$ as the height above some chosen ‘reference level’ (say the floor). And then use GPE = mgh.

Suppose the system is in equilibrium but suddenly the mass becomes detached from the spring.

When the spring and mass finally come to rest on the floor (after bouncing around a bit) the ‘mechanical energy’ has been dissipated – causing heating.

(If this occurred in an isolated system and we knew the overall heat capacity, we could even find the temperature-rise.)

In the final state, GPE = EPE = 0 (using the usual conventions). So the required mechanical energy is simply the sum of the initial GPE and the initial EPE. (This, I think, is a description of what @haruspex referred to as ‘readily available mechanical energy’.). The initial mechanical energy can easily be expressed in terms of $M, g, h$ and $k$ only.

 

[kuruman]

(say the floor).

Why say the floor and not a chair that is higher than the floor? We are told that "the mass is now at height h". OK, but if it is recoverable energy that we are talking about, the height of the mass's final resting place is needed to figure out this recoverable energy.

But we don't care what the energy transformations were to reach the stated arrangement, only what we can then extract.

Starting at the stated arrangement, how do we know how much energy we can extract without specifying the final arrangement? Potential energy depends on two end points and the amount of energy extracted depends on the knowledge of both. If the final arrangement is not specified but the zeroes of potential energies are, then it can reasonably be assumed that the final arrangement is where the system is returned to zero potential energy.

 

[Steve4Physics]

Why say the floor and not a chair that is higher than the floor? We are told that "the mass is now at height h". OK, but if it is recoverable energy that we are talking about, the height of the mass's final resting place is needed to figure out this recoverable energy.

The real problem is that the question is badly posed.

However, it appears to be set at an introductory level and on that basis, it seems reasonable to assume that $mgh$ is intended to be the required GPE-contribution to the mechanical energy.

The question can then be answered with that assumption.

Edited.

 

[nasu]

Suppose the system is in equilibrium but suddenly the mass becomes detached from the spring.

When the spring and mass finally come to rest on the floor (after bouncing around a bit) the ‘mechanical energy’ has been dissipated – causing heating.

(If this occurred in an isolated system and we knew the overall heat capacity, we could even find the temperature-rise.)

In the final state, GPE = EPE = 0 (using the usual conventions). So the required mechanical energy is simply the sum of the initial GPE and the initial EPE. (This, I think, is a description of what @haruspex referred to as ‘readily available mechanical energy’.). The initial mechanical energy can easily be expressed in terms of $M, g, h$ and $k$ only.

Why is the spring resting on the floor if the mass is detached from the spring?
If the mass is detached, the spring energy remains with the spring which I suppose will relax. If it's assumed to be a masseless spring, it will have no inertia to oscillate around equilibrium. "Recovering" the energy of an ideal spring is problematic once is detached from the massive part of the system.
I don't see any way to "recover" (or make available) both the gravitational and elastic potential energy of this system with ideal components.
It's likely that the author did not give it too much tought and just intended for the student to add the gravitational and elastic PEs with the standard reference levels used in intro textbooks.

 

[haruspex]

I don't see any way to "recover" (or make available) both the gravitational and elastic potential energy of this system with ideal components.

First, you fix the end of the spring attached to the mass in place. Next, you allow the mass to descend to the floor doing work mgh in the process. Then you allow the spring to relax, doing work $\frac 12kx^2$.
Or do it in the other order, fixing the mass in place first.

 

[kuruman]

doing work mgh

doing work $\frac1 2kx^2.$

Doing work on what?

the system should include mass, spring and Earth.

 

[jbriggs444]

Doing work on what?

Stuff outside the system.

For instance, the falling mass could be attached to a piston and cylinder arrangement so that a pressurized fluid could flow past a turbine and an attatched electrical generator.

I see no problem carving an enclave within the closed system, drawing interfaces for the free end of the spring, the bottom of the mass and an anchor point on the earth. Then we can clearly identify energy flows out of the closed system and into the enclave.

 

[haruspex]

Doing work on what?

the system should include mass, spring and Earth. It does not have to be limited to those.

 

[kuruman]

All that is fine but does not address the original question " Calculate the amount of mechanical energy if the mass is at rest." I can understand that a certain amount of Joules can be recovered by all sorts of processes that cross the boundaries of the system.

If these processes conserve mechanical energy, the initial mechanical energy does not change. The "recovered" energy is energy converted into a different form of mechanical energy by conservative forces. We can find that number but how does that help us find the total that is conserved?

I take $8.53 out of my left pocket and put it in my right pocket. I then ask how much money I have in my two pockets together. The reasonable answer is, the total amount is unknown but you have at least $8.53 in your pockets.

If the processes of energy conversion are not conservative, then it becomes worse. My right pocket has a hole and one cannot even guarantee that I have at least $8.53.