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A mass m is placed on a vertical spring and allowed to reach equilibrium, whose level is ##e_0## below the original level of the spring before placing the mass. It is then pushed downwards such that it is now a distance ##x_0## below the equilibrium level. What is the maximum height d from the level of release it can reach?

Method 1

By conservation of energy, loss in elastic potential energy (EPE) = gain in gravitational potential energy (GPE).

$$\frac{1}{2}k(e_0+x_0)^2=mgd$$

$$d=\frac{k(e_0+x_0)^2}{2mg}$$

Method 2

After release, the mass moves upwards and loses contact with the spring at a level distance x above the equilibrium level. At the point where contact is lost, ##a=g=\omega^2x##, where ##\omega=\sqrt{\frac{k}{m}}##, since it is in simple harmonic motion.

And velocity $$v=\omega\sqrt{x_0^2-x^2}$$

By conservation of energy, loss in kinetic energy (KE) = gain in gravitational potential energy.

##\frac{1}{2}mv^2=mgh##, where h is the distance from the point where contact is lost to the maximum point. Making h the subject and substituting v from the equation above and ##x=\frac{g}{\omega^2}##, $$h=\frac{\omega^2x_0^2}{2g}-\frac{g}{2\omega^2}$$

$$d=h+x+x_0=\frac{\omega^2x_0^2}{2g}+\frac{g}{2\omega^2}+x_0=\frac{kx_0^2}{2mg}+\frac{mg}{2k}+x_0$$

It can be shown that the two answers are the same. My question is just before contact is lost, the mass has EPE, KE and GPE, but just after contact is lost, it only has KE and GPE, so where does the EPE go to? EPE isn't included in the calculation of h in method 2 (in the step where conservation of energy is used). So it seems like the mass loses some energy (equal to the EPE) after it loses contact. So I would expect method 2's answer to be smaller than method 1's, where no such loss in energy is used.

A mass m is placed on a vertical spring and allowed to reach equilibrium, whose level is ##e_0## below the original level of the spring before placing the mass. It is then pushed downwards such that it is now a distance ##x_0## below the equilibrium level. What is the maximum height d from the level of release it can reach?

Method 1

By conservation of energy, loss in elastic potential energy (EPE) = gain in gravitational potential energy (GPE).

$$\frac{1}{2}k(e_0+x_0)^2=mgd$$

$$d=\frac{k(e_0+x_0)^2}{2mg}$$

Method 2

After release, the mass moves upwards and loses contact with the spring at a level distance x above the equilibrium level. At the point where contact is lost, ##a=g=\omega^2x##, where ##\omega=\sqrt{\frac{k}{m}}##, since it is in simple harmonic motion.

And velocity $$v=\omega\sqrt{x_0^2-x^2}$$

By conservation of energy, loss in kinetic energy (KE) = gain in gravitational potential energy.

##\frac{1}{2}mv^2=mgh##, where h is the distance from the point where contact is lost to the maximum point. Making h the subject and substituting v from the equation above and ##x=\frac{g}{\omega^2}##, $$h=\frac{\omega^2x_0^2}{2g}-\frac{g}{2\omega^2}$$

$$d=h+x+x_0=\frac{\omega^2x_0^2}{2g}+\frac{g}{2\omega^2}+x_0=\frac{kx_0^2}{2mg}+\frac{mg}{2k}+x_0$$

It can be shown that the two answers are the same. My question is just before contact is lost, the mass has EPE, KE and GPE, but just after contact is lost, it only has KE and GPE, so where does the EPE go to? EPE isn't included in the calculation of h in method 2 (in the step where conservation of energy is used). So it seems like the mass loses some energy (equal to the EPE) after it loses contact. So I would expect method 2's answer to be smaller than method 1's, where no such loss in energy is used.

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