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I thought this would be a particularly simple problem but it is turning out to be quite the opposite. I am sure I am doing a very simple mistake.

The problem statement is that there is a mass which is just barely kept on the spring (help by the force equal to the weight of the spring) such that the spring is uncompressed. As soon as I let the object go, it will compress the spring and come to rest at some height. The energy stored in the spring will be equal to the difference in the potential energy at the two height of the spring (compressed and uncompressed),

$$mgh_i -mgh_f= 1/2 kx^2$$ Furthermore, the compression of the spring would just be the difference in the $h_f -h_i$. This gives me two roots for compression,

$$ x=0 $$ And the second would be, $$ mg=1/2kx $$ But then by this, $kx=2mg$ and $kx$ is force by Hooks law, but then does it means that the force compressing the spring is twice the weight of the object? That sounds odd. I was expecting I would just recover Hooks law but I guess I am doing something wrong here. But I am not sure what that is.

Any help will be much appreciated.

P.S. This is not a homework problem. We are designing a project for our school.

Thanks for your time.

Peter Shor
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Shaz
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    Hint: What happens when you let the object go, assuming no friction, is that it starts bouncing up and down on the spring and never stops. So if it stopped, some of the energy was dissipated by friction. – Peter Shor Jul 04 '16 at 11:28
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    We treat homework-like problems the same way as homework here, which means we try to give hints and not complete answers. – Peter Shor Jul 04 '16 at 11:31
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    As a further hint to Peter's, if you compute the work done lowering the mass to equilibrium (so that it doesn't oscillate) then you will get your missing $1/2kx$ of energy. – lemon Jul 04 '16 at 11:52
  • Ok, first, as I said before, its not a homework problem. The calculation is part of a equipment we are building for our school.

    The mass does oscillate a little in practise. It comes up in the data of the experiment. But from a mathematical point of view that should not be important because that is a parameter of the design and you could argue for such a case, mathematically, that it does not oscillate.

     – Shaz Jul 05 '16 at 05:11
  • No you can't argue that it does not oscillate. If there is no damping, all potential gravitational energy must become elastic energy - and you get the result you calculated. But then the spring must oscillate indefinitely. If you allow for energy dissipation, the compression will be less. In the limit where you very gradually release the spring, it will come to rest at the equilibrium position - in that case gravity did some work on you, and less work on the spring. – Floris Jul 07 '16 at 14:20
  • Ok. So I will tell you guys a little of what we were trying to do. We are trying to make a force sensor: If you punch it, it should be able to tell you how much force was used in the punch. The idea is that the punch will compress the spring and if I know how much the spring is compressed I can tell the force of punch just by Hooks law. This particular problem occurred when I was trying to understand what happens when I drop a weight on the sensor from different heights. – Shaz Jul 11 '16 at 03:48

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When the spring reaches maximum compression, the mass is instantaneously at rest but it is not in static equilibrium. The net force on it is not zero : $kx \ne mg$. Like a pendulum at the end of each swing, there is a net force on the mass causing it to accelerate towards the equilibrium position - at which the net force on it is then (for an instant) zero. Your calculation is correct. It is your interpretaion of the result which is at fault.

At the lowest point the compression force in the spring is $kx=2mg$, acting upwards on the mass, while gravity is still pulling down with force $mg$. There is a net force of $mg$ acting upwards.

sammy gerbil
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  • Hi, thanks for your answer. There is something else as well. The energy equation comes from calculating the work done. Since the force is conservative, then $F=-\nabla E$ (this equation is vector, the symbols were coming all screwed up). Putting in the energy equation $E=1/2kx^2$ you do get Hooks law back. Then why am I not getting it otherwise.

    The spring is pushing it upwards and weight is pushing it downwards. Both should balance out and the force should be zero i.e. F+W=0.

     – Shaz Jul 05 '16 at 15:05
  • @Shaz : No, the forces do not cancel out. The system is not in static equilibrium at maximum compression. The mass moves back up and oscillates. That is the point of my answer. See comments by Peter Shor and lemon. – sammy gerbil Jul 05 '16 at 15:37
  • Ok, so I do get the idea about mass oscillating. And that I am missing a factor in the energy equation. I tried something else as well. To get a better idea of what I am missing, I tried adding a function, say $D$, such that I dont get this problem i.e. I tired $mgh_i=1/2kx^2 +mgh_f +D$ and I fixed D such that I necessarily get mg=kx. It turned out that $D$ should be something like potential energy term, $D=(h_i-h_f)1/2(mg)$. I am not exactly sure what this means. – Shaz Jul 11 '16 at 03:44