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I have reached a confusion. We know $$F=-kx$$ in a spring. If an external mass $m$ is suspended to a spring and the spring extends by a length $x$ and the mass comes to rest,

external force, $~mg = \text{restoring force of spring} ~~kx$ ( talking about the magnitude only here)

Again, at the stopping point, work done by external force will me $mgx$ and potential energy gained by spring will be $$\frac{kx^2}2$$ These terms must be equal as the system is in rest

So $$k\frac{x\cdot x}2 = mgx$$ Cancelling out $x$, we get $$\frac{kx}2 = mg$$ These contradicts the original statement that $mg=kx$

Please help me. I am confused.

user8718165
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c0mpleX
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  • Also similar: https://physics.stackexchange.com/questions/127002/wrong-calculation-of-work-done-on-a-spring-how-is-it-wrong – BowlOfRed Apr 23 '19 at 15:06

3 Answers3

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If you hold the mass at $x=0$ and then release it, it does not immediately come to rest when $x = mg/k$. It performs simple harmonic motion, oscillating between $x=0$ and $x=2mg/k$.

In real life it eventually stops because the damping forces take away energy from the system as it oscillates.

If you haven't studied damped oscillations yet, you can see your mistake by supposing that you lower the mass slowly from $x=0$ until it is in equilibrium. As you do that, the force you apply to the mass to support it reduces linearly from $mg$ to $0$, so the average force you apply is $mg/2$ and the amount of work that you do while lowering the mass (force time distance) is $-mgx/2$.

So half of the work done by gravity ($mgx$) is actually done on you, as you slowly lower the mass. The other half is stored as the potential energy in the stretched spring.

alephzero
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This system is not absent from external forces ,and that is why you cannot apply the conservation of energy directly. The spring force and gravity are the forces that act on the system.

But nevertheless, this executes SHM between the release point (assuming of course that the spring is in the natural length) and the extrema point when $v=0$, which will be at $x=2mg/k$.

Karthik
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Assume a spring fixed to a ceiling, with a weight attached to the bottom of the spring. If you push up on the weight and hold it against the bottom of the spring such that the spring is initially unstretched and uncompressed, the total energy of the mass/spring system will be $mgh$, where $h$ is measured from the point of maximum spring stretch (currently unknown).

When you let go of the weight, it will fall until all of the gravitational potential energy of the system is converted into spring potential energy. To determine how far the weight will fall, solve the equation $mgh=1/2kx^2$ for $x$, where $x$ is also equal to $h$.

Note that this is the lowest point that the weight can reach, and as soon as the weight gets to this point, the spring will pull the weight up, but the weight will not get quite as high as the release point due to air drag, and other dissipative forces (e.g., a small amount of friction in the spring). The weight will continue to oscillate up and down, with a slowly decreasing amplitude, until it comes to rest somewhat below the release point. At this point, you will find that $mg=kx$. This means that your analysis above involves two completely different situations: one situation describes the maximum stretch in the spring, and the other situation describes the new equilibrium position of the spring when it stops oscillating.

David White
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