If string is stretched just by weight, where does the gravitational potential energy?



Answered question


If a spring is stretched by a weight of mass m, (so the extension is Δx) then
k = W Δ x = m g Δ x . So k Δ x = m g
When the Spring is stretched by distance Δ x (by the weight) then it looses gravitational potential enegry. ( Δ G P E = m g Δ x)
But, when we calculate the change in elastic potential energy, We get
U = 1 2 k ( Δ x ) 2 . Since k Δ x = m g, U = 1 2 m g Δ x
At, equilibrium we have no kinetic energy E k = 0
Doesn't that violate Energy Conservation?

Answer & Explanation

Ricky Lamb

Ricky Lamb

Beginner2022-09-07Added 7 answers

If you attach the weight and let it drop, it will fall and gain kinetic energy, overshoot the equilibrium point, slow down, change direction, and then cycle over and over again. In order for it to ever reach equilibrium, the energy must dissipate as heat via air resistance, friction in the spring, etc.
Chasity Kane

Chasity Kane

Beginner2022-09-08Added 2 answers

This is a subtle matter. The block needs to move quasi-statically to keep its kinetic energy zero, and for that reason we would need an external force F. This force will also do negative work on the block, reducing its change in GPE by half, which is stored in the spring. If the process was not quasistatic, the block would have a velocity after travelling some distance, and your question is answered, since you need to take the kinetic energy then as well. Note that F is not a constant force, because it has to counter the effect of gravity as well as the changing spring force.

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