Spring constant, vertical ball launch speed.

[SteelDirigibl]

Ok I have most of this solved but can't get the third part right.

Homework Statement

A spring of spring constant 128N/m is compressed a distance of 2.0m from its equilibrium position, and used to project a ball of mass 4.0 kg directly upwards. Neglect air resistance.

1. What is the potential energy of the spring in its compressed position? 256 J
2. To what maximum height above its initial (compressed) position does the ball reach? 6.53m

3. Earlier, just when the spring is returned to its equilibrium position, as the ball was moving upwards, how fast was the ball moving? undetermined?

Homework Equations

KE=1/2*mv²
PE=mgh
U_s=1/2kx²

The Attempt at a Solution

The first thing I tried was simply using the previously calculated energy, 256J, and solving for velocity in the KE equation. This gets 11.3 m/s using mass of the ball, but I'm sure I'm supposed to account for some other motion of the spring, because it doesn't transfer all of that energy to the ball. The correct answer is 9.4 m/s. Where do I go from here?

 

[Doc Al]

Don't assume that all of the spring PE goes into KE--the ball also rises. Use conservation of total mechanical energy.

 

[SteelDirigibl]

Ah I get it.

The ball rises 2.0m at that point. PE of the ball is 2*4*9.8=78.4

256-78.4=177.6

1/2*m*v^2=177.6

177.6*2/4=88.8
88.8^.5=9.42 m/s

Thanks a lot! it looks like I just need to think a little more about what I'm doing. Got my final tomorrow morning.