Elastic Potential Energy and velocity of spring

[euphoriax]

Homework Statement

A spring, having a force constant of 6.0x10² N/m, is held in a vertical position and compressed 0.30m. A 5.0 kg mass is then placed on top of the spring. THe mass is then releases. Neglecting air resistance and the mass of the spring, calculate:
a) the velocity of the mass when it has moved up 0.40m from the compressed position of the spring.
b) the velocity of the mass when it has moved up 0.20m from the compressed position of the spring.

Homework Equations

(E_pe + E_k)i +/- W = (E_pe +E_k)_f

The Attempt at a Solution

a) h= 0.4m, v_f=?

(E_pe + E_k)i +/- W = (E_pe +E_k)_f
0.5kx²=0.5mv_f² +mgh
kx²=mv_f2 +2mgh
v_f= √((kx²-2mgh)/m)
= √( ((600)(0.3)²-2(5)(9.81)(0.4))/ 5)
v_f = 1.7 m/s [up]


b) h=0.20m
0.5kx²=0.5mvf² +mgh
kx²= mv²+2mgh
v_f= √((kx²-2mgh)/m)
= √( ((600)(0.3)²-2(5)(9.81)(0.2))/5 )
v_f= 2.6 m/s [up]

the answer for b) is 2.4 m/s [up], so I'm off by 0.2m/s. Can anybody explain to me what was wrong? I appreciate it

 

[rl.bhat]

In the second case only part of the compressed energy is used to push the block upward. So x should be 0.2 m, not 0.3 m.

 

[euphoriax]

But before the mass is released it's still being compressed 0.3m, and then after it moves up 0.2m.

I tried that anyway, and it wasn't the correct answer...

 

[PhanthomJay]

But you are neglecting the fact that the spring still has EPE at 0.2 m up from the initial compressed position . You must include that EPE of the spring on the right side of the conservation of energy equation.