Problem with energy conservation of mgh and 1/2 kx^2 (spring)by nickypoo Tags: 1 or 2, conservation, energy, spring 

#1
May313, 12:46 AM

P: 8

So I feel really stupid asking this, because this is a very elementary physics problem and I'm well past this level of physics, but I don't understand what's going on here.
Say we have a vertical spring with spring constant 100N/m. A 100N mass is placed on top of it. This will cause the spring to compress by 1 meter by Hooke's law. Now the issue I have is with energy conservation. The work done on the spring in terms of the mass is mgh = (100N)(1m) = 100J. In terms of work done on the spring, we have Energy stored in spring U = 1/2kx^2 = (0.5)(100N/m)(1m^2) = 50J. So where does this energy discrepancy come from? I know that energy is conserved, so the problem lies in my methodology. Am I missing a component of velocity that the mass will have? I was assuming that if the mass was added to the spring slow enough, it wouldn't oscillate. 



#2
May313, 01:18 AM

P: 865

How do you add to the mass to the spring slowly? This entails a third party doing work on the mass doesn't it? Something that will rob the kinetic energy from the system. A viscous fluid or a hand perhaps? I think that is where the issue may lie. Cool question!




#3
May313, 03:05 AM

P: 2,861

Lets say you suspended the mass just above the spring using a rope. When you cut the rope the mass will fall compressing the spring.
The key point you missed is that when it has fallen 1 meter (eg when it has compressed the spring by 1 meter) it will still have velocity and hence KE. The equation will be.. Gravitational PE + Spring PE + KE =0 or mgh + (1/2)kx^{2} + (1/2)mv^{2} = 0 The mass will bounce up and down trading PE and KE back and forth. It won't come to rest at x=1m unless/until something removes energy from the system (eg your hand or air resistance). 



#4
May313, 03:09 AM

P: 8

Problem with energy conservation of mgh and 1/2 kx^2 (spring)
@ ModusPwnd: Yes I think you're right. The mass would do work on an external system that slows it down. I guess my conceptions are not fully developed.
The spring will have maximum compression when the kinetic energy of the mass is zero. All 100J of gravitational kinetic energy will be converted into potential energy of the spring and it will compress according to U=1/2kx^2 before it springs back up, oscillating around equilibrium. That must have been my problem, thinking of a static load on the spring vs. what happens to gravitational potential energy and the work being done on the spring by the descending mass. @CWatters: I wrote my comment above as you were posting yours. Thank you for expressing the circumstances mathematically. It's clear now where I went wrong. What an eyeopener to how much I've forgotten. I am in the 3rd semester of Physics now, and got an A in 1st semester Physics that covers Classical Mechanics. It's distressing to me that I couldn't discern the relationships involved with this. 



#5
May313, 10:52 AM

C. Spirit
Sci Advisor
Thanks
P: 4,941

It might be instructive to actually solve for the motion assuming no air resistance etc. Taking the reference to be the unstretched length of the spring, we can write the Lagrangian for the system as ##L = \frac{1}{2}m\dot{y}^{2}+ mgy  \frac{1}{2}ky^{2}##. Lagrange's equations then give us ##\ddot{y}+ \frac{k}{m}y = g ##.
Note that this is just a forced oscillator and the solution is (after using the initial conditions ##y(0) = \dot{y}(0) = 0##) ##y(t) = \frac{g}{k}\cos\omega t + \frac{g}{k}## where ##\omega## is the natural frequency of oscillation. Above you solved for the displacement ##y(t_0)## when ##\ddot{y}_{t_0} = 0## (i.e. when ##y(t_0) = \frac{mg}{k}##) which implies ##\dot{y}_{t_0}## is a maximum. In particular, it is ##\dot{y}_{t_0} = \frac{g}{k}\omega##. You assumed it would have no kinetic energy at ##t_0## which as you can see is false. If you want however, you can add a damping term to the equations of motion (possibly due to air resistance), solve the equations of motion, and adjust the damping constant so that the mass does come to rest when the spring has been displaced 1 meter from the unstretched length. 



#6
May413, 03:09 AM

P: 102

First off, you have “100N mass” which is wrong since mass is not measured in kilograms, not in Newtons. It’s force that is measured in Newtons and that force is the force of gravity I think you meant, “An object that weighs 100N is put on top of it.” Now please excuse me while I vent a pet peeve of mine: It’s not quite correct to say “A 100 kg mass is placed on top of it.” since mass is a dynamical quantity and not a descriptive quantity. One should say instead “An object whose mass is XXX kg which therefore weighs 100N is placed on top of it.” Now, on to bigger and better things. :) Notice the details of how this is done. Visualize it in our mind. How are you doing this? Did you attach the weight to the unloaded spring and then let go of it? If you did then the sum of kinetic energy and potential energy of the spring and gravitational potential must be taken into account since the weight would be oscillating up and down in sinusoidal motion. As you lower the object the gravitational force acting on the weight causes a downward force. It is this force which is compressing the spring and the force that is doing work. Since you’re lowering it at constant speed your hand must be supporting it by applying an upward force equal to the sum of the force exerted by the spring plus the force of yoru hand on the weight. It is your hand that is doing an amount of work that is negative. Since the kinetic energy is zero when it comes to rest at the bottom then the total work done on the system is zero. The potential energy stored in the spring comes from two sources. There is energy lost by the system because the system is doing work on your hand. When all these things are taken into account you will find that energy is conserved. Please take note that it’s the energy of a closed system that is conserved and since your hand is acting on the system the system is not closed. 



#7
May513, 12:12 AM

P: 7

I'm thinking like this:
At the top, there is only one force W= mg = 100 N. Clearly its going to accelerate down. Before it reaches 1 metre, mg is going to be greater than F = kx, so its still accelerating downwards and gaining velocity. At the x = 1 m point, there are two forces, W = 100 N down, and F = kx = 100 N up. For a net force of 0 N. Since there are no forces on the object, there is no acceleration at this point, and it corresponds to the centre of SHM oscillations. Half of the 100 N of GPE will have gone into compressing the spring and the other half into the KE of the object. This is the maximum of velocity, since once the block moves below this point, the force kx upwards will exceed mg down, so it's being accelerated up, i.e. its slowing its downward velocity. When it finally gets to the bottom, x will be 2 m, and 1/2 kx[itex]^{2}[/itex] is going to be 200 J. The GPE lost will be 200 J, there is no KE, and the energy is in balance now with no v or KE. BUT now there is a force upwards, of 200 J (from kx) minus 100 J (from mg) = 100 N upwards. And the object will accelerate up and the cycle will keep repeating in SHM oscillations. I was trying to show this more clearly by working out what the KE would be in the middle at x=1m, but then I got stuck figuring out how to calculate v in terms of x. Anyone know how to do this? 



#8
Jun1613, 02:13 AM

P: 12

At the point where the object has descended 1m, the spring will be exerting a force of 100N, as will the force of gravity. However, this says only says that the velocity is not changing, not the position. Before reaching the 1m point, the force exerted by the spring would have to have been less than 100N, so it would have acquired a nonzero descending velocity by the time it descends one meter. The remaining 50J are the kinetic energy that it has when it has fallen that much.



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