Calculate this Spring's Potential Energy

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  • #26
PhanthomJay
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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.
 
  • #27
kuruman
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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
$$mgy=\frac{1}{2}ky^2-W_{hand}.$$
 
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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"
 
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@kuruman @PhanthomJay

Thank you for your answers and efforts.
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 :)
 

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