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A 10-g mass is attached to the end of an unstressed, light, vertical spring (k = 49 N/m) and then dropped. Answer the following questions by considering the potential energy due to the spring plus the potential energy due to gravity, i.e. measure distances from the equilibrium position of the spring with no mass attached. (a) What is the maximum speed of the falling mass? (b) How far does the mass drop before coming to rest momentarily? (c) Repeat (a) and (b), but answer the questions by considering the potential energy of the spring with the mass attached, i.e. measure distances from the equilibrium position of the spring with the mass attached.

As it turns out, the maximum velocity is 14 cm/s and the distance it drops before coming to a momentary rest is 4 mm.

In solving for the distance, I went the route he took in class which is as follows:

Net Force = 0

-Fspring - Fgravity = 0

-kx - mg = 0

Assuming g = 9.8, when I plug in the rest of the values I get a -.002 m for x, the negative of course falling in line with assuming that the initial position is 0 and it moves downward. This of course does not agree with the 4 mm answer I should be coming to.

As far as maximum velocity is concerned, well there is an initial potential due to gravity as well as an initial potential due to the spring (which can be eliminated by setting the initial position at 0, right?) and a final potential due to the spring as well as a final kinetic energy.

All of this is according to part c as far as I can tell, as the mass of the block is considered in the problem. I think I kind of understand it, though by no means do I have a grasp on it (as you can see). But this also leaves out solving any of it without considering the mass to be a part of the problem, as is asked in parts a and b.

Any and all help would be much appreciated!