If a 2.0-kg mass is now attached to the end of the spring, and the mass is allowed to fall, where will the end of the spring line up with the ruler marks when the mass is at its lowest position?

**Answer 1**

I agree with your answer if we ignore the transients.

The question is worded such, using the word “fall”, that they may be looking for the maximum spring stretch as it bounces without friction as in a simple harmonic oscillator.

To do this, set the change in potential energy of the mass equal to the maximum spring potential energy

PE = PS

mgh = ½kx²

then recognize that h and x represent the same distance

mgh = ½kh²

mg = ½kh

h = 2mg/k

h = 2(2)(9.8)/(53)

h = 0.7396

notice that this is exactly twice the distance of stretch as compared to if you lower the mass slowly to its new neutral position. (which is what your original analysis assumed)

19 + 74 = 93 cm

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**Answer 2**

The trick is in the “lowest point”.

It is not when the mass comes to rest.

It’s when it passes its equilibrium point (which is at 56 cm), and overshoots, slowing down. When V = 0, that’s the lowest position, and then it will bungee back up.

So first figure the Ep of the spring (work done by the spring) at equilibrium.

Ep = 1/2*kx^2 = 3.62 J

Then the Ek of the mass at that point is the difference of mgh and W.

2*9.8*0.37 – 3.62 = 3.62 J

Now you’re simply finding the point at which the work done will cancel that out, i.e. where Ek = 0 and W = 3.62 J.

In other words, the total work from start will be 7.25 J.

W = 1/2*kx^2

x = 0.523 m

So mark at ruler = 71 cm.

Best wishes!

**Answer 3**

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