0
$\begingroup$

I am having quite difficult time to understand the meaning of total potential energy on spring.

All references I read say that the total potential energy on particle attached to spring is:

 total potential energy (U)=elastic potential energy (from spring) + potential energy from external force

The work done on particle due to system of forces is equal the difference in potential energy between point A and point B.

If we chose point A to be the datum with 0 potential energy, then the total work done on the particle to move it from point A to point B is equal to the potential energy of the particle at point B.

 U (Total potential energy) = W (Total work) = Ws (word done be spring force) + We (work done by external force)

enter image description here

In this picture, it is well know that the spring force is a function of the displacement x so the work done by spring force on the red particle is

dws=-Fs.dx=k.x dx
ws=-1/2kx^2

Now the problem is in the work done on the particle by the external force F. References I read say that the work done by the external force is

we=F.x 

because the force is independent of displacement x.

My question is as follows:

At the initial stat, there is no force F and no spring force Fs. Once we apply the external force one the particle the work start to be exerted on the particle.

we normally do that gradually and the external force F should be larger than the spring force Fs in order to move the particle to the left.

Through distance x, external force should be larger than the spring force and then at some point it should be adjusted to make the total force on the particle equal zero at final position (with zero velocity).

so I thing the work done be external force should be some thing like this:

we=Integrat(F(x,t)dx)

Why the external force is considered independent of the displacement x and then the work done by external force is calculated as:

we=F.x
$\endgroup$
  • 1
    $\begingroup$ I'm still confused about what you're asking... The external force is considered independent because it's an imposed force outside the system. We're making it independent. That's the boundary condition of the system. $\endgroup$ – tpg2114 Aug 28 '14 at 2:44
  • $\begingroup$ let us assume you pull the particle attached to spring (using your hand) to some distance, do you think the force you exert is constant all the time? $\endgroup$ – Algohi Aug 28 '14 at 2:49
  • $\begingroup$ You're talking about two different things. In real life, of course not. In an idealized text-book problem, yes. You're trying to solve the latter but thinking of it like the former. $\endgroup$ – tpg2114 Aug 28 '14 at 2:51
  • $\begingroup$ In the derivation, the external force is equal to the spring force and the spring is compressed slowly. That is called quasi-static conditions. $\endgroup$ – Spirko Dec 8 '15 at 5:35
  • $\begingroup$ Your question has a lot of unclear statements. That might be one source of your confusion. I almost stopped reading after the first statement in gray because it's very vague. I did stop reading half-way through. I was lost. That first statement in gray: total potential energy of what? What is your system? Potential energy from eternal force: What do you mean by that? Are you including the device that creates the force in your system? That's the only way I can understand that. If your system is the spring, PE is elastic and nothing else. $\endgroup$ – garyp Feb 13 '16 at 13:26
1
$\begingroup$

Are you wondering why your two equations $W=\frac{1}{2}kx^2$ and $W=Fx$ don't match? That's because the latter equation, $W=Fx$, is only true for a constant force. The more general expression is $W=\int \vec F \cdot d \vec x$, similar to what you wrote in differential form $dW=F\,dx$.

The expression with $\frac{1}{2}$ in it is correct. The expression with just $Fx$ is only correct for constant forces, which spring forces are not.

$\endgroup$
  • $\begingroup$ I am not talking about the spring force. I know it is not constant through the displacement. my concern is the external force F. it is not constant when you pull the spring. apparently you should gradual increase F from 0 to final value. this is what confuse me. even with the gravity force, when you hang a particle, you don't let go once. you slowly let it go which means that the net external force (constant gravity force and your hand force) is not constant through the displacement. $\endgroup$ – Algohi Aug 28 '14 at 2:15
  • 1
    $\begingroup$ Can you add that description or clarify it in your question? As is, it's a bit hard to tell what you're asking. $\endgroup$ – BMS Aug 28 '14 at 2:26
  • 1
    $\begingroup$ @BMS If you suspend a mass from a light spring and let it go, then you have a case where the force on the spring is in fact nearly constant ($mg$). The fact that there are other cases where the force is not constant (pulling a string by hand) should not matter. $\endgroup$ – BowlOfRed May 5 '15 at 7:43
0
$\begingroup$

I think your confusion is between quasi-static extension of spring and extension of spring by a constant force.

Here are some pointers:

  1. The potential energy of the spring $U_s$ is always going to be $\frac{1}{2} k\Delta x^2$ for a displacement $\Delta x$ from its "relaxed" state which we can set as the zero potential energy state.

  2. The work done in getting the spring to be displaced by $\Delta x$ is going to be at least that amount --it will be more if you end up imparting a velocity to the end point. If you do it in a manner where the velocity of the end point remains zero (i.e., you stretch the spring quasi-statically) then this work will equal the amount of potential energy stored in the spring (as given by #1). Note that, you can not quasi-statically stretch the spring with a constant force --as you yourself are wondering in the question. In this case, $\int_0^{\Delta x} F(x) dx$ = $\frac{1}{2} k\Delta x^2$ (as the kinetic energy of the mass at the end of the spring is $0$).

  3. If the force applied to the end of the spring is constant then you will end up giving some kinetic energy to the mass attached at the end of the spring. How much? Well, the work you did is $F \Delta x$ and the amount that went in the spring is $\frac{1}{2} k\Delta x^2$. Therefore, the kinetic energy of the mass attached at the end is $F \Delta x - \frac{1}{2} k\Delta x^2$.

$\endgroup$
  • $\begingroup$ And it would be better to call the last F in part (3) a different notation such as mg or something else. I think this is the main source of confusion in the understanding the problem. You can use W=F.d only if you know that F is fixed and you are moving along a straight line which is the case for lifting to lowering an object in the gravitational field of planet Earth. This definition of work is deduced from the most general case where it is the integral mentioned in part(2). So, don't confuse between the most general definition and when it is reduced to a multiplication for special cases. $\endgroup$ – Benjamin Mar 14 '16 at 20:51
0
$\begingroup$

I think I got the same question as you. I have got my answer:

In this pulling spring process (from A to B), We use an assumption that is not clearly stated: The external force F is constant, but the internal force in the spring is gradually increased. This can't be true in the real world, but here we use this assumption and give it a concept "quasi-static". That's why we have internal strain energy: 1/2 * k * x^2, and potential energy of the external force: - F * x. And finally, we get the total potential energy of this system under the quasi-static condition: 1/2 * k * x^2 - F * x

$\endgroup$
  • $\begingroup$ We do have mathjax enabled on this site to have LaTeX-like equations. Search 'notation ' in help center for for details. $\endgroup$ – Kyle Kanos Jun 17 at 16:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.